There are some problems with that question. One of them is that it doesn't state how the passwords are chosen but I think the most logical approach is to assume the passwords are chosen randomly but satisfying the respective conditions so I'll use that convention for my answer. Note that Randall's comic clearly doesn't share this assumption but the question didn't specify which way a password is chosen so I reckon we can go for the best which is possible and that's choosing a password randomly. Furthermore, the test probably isn't based on Randall's comic.

The key pace of option b is quite easy to calculate if we assume the English alphabet is used. Yeah, more assumptions, I know. But since the test appears to be in English and not very tricky, I think we can make that assumption.

There are 26 lower-case letters in the English alphabet and just as much upper-case letters, making 52 in total. So there are 52^10 ≈ 1.45*10^17 elements in the key space of option b.

Option c is way less specific than option b. However, since we assumed that the English alphabet is used – which is in favor of option c – we may also assume that only ascii is used for the special characters – which is in favor of option b. Really, if we assumed more special characters than ascii has, we got to assume more letters than are in ascii since ä arguably is a letter in German. That makes the key space of option b even bigger compared to the one of option c.

The best we can do for option c if we restrict ourselves to the ascii alphabet is to use every printable character (excluding the blank) in our alphabet (note: different use of the word "alphabet"). That's 94 characters, giving option c a key space of 94^7 ≈ 6.48*10^13 elements.

Since one of our assumptions to tackle the question is that the password is chosen randomly witch the respective restrictions and that rule is equal to choosing a password randomly from the respective key space, a password chosen using option b is arguably harder to guess since there are several orders of magnitude more options to try when cracking the password.

In fact, if we assume the costs of cracking a password via brute force to be approximately linear to the size of the key space, cracking a password chosen via option b is 52^10/(94^7) ≈ 2'229 times as hard as cracking one chosen via option c, clearly showing that the allegedly correct answer to this question is wrong.

There are some problems with that question. One of them is that it doesn't state how the passwords are chosen but I think the most logical approach is to assume the passwords are chosen randomly but satisfying the respective conditions so I'll use that convention for my answer. Note that Randall's comic clearly doesn't share this assumption but the question didn't specify which way a password is chosen so I reckon we can go for the best which is possible and that's choosing a password randomly. Furthermore, the test probably isn't based on Randall's comic.

The key pace of option b is quite easy to calculate if we assume the English alphabet is used. Yeah, more assumptions, I know. But since the test appears to be in English and not very tricky, I think we can make that assumption.

There are 26 lower-case letters in the English alphabet and just as much upper-case letters, making 52 in total. So there are 52^10 ≈ 1.45*10^17 elements in the key space of option b.

Option c is way less specific than option b. However, since we assumed that the English alphabet is used – which is in favor of option c – we may also assume that only ascii is used for the special characters – which is in favor of option b. Really, if we assumed more special characters than ascii has, we got to assume more letters than are in ascii since ä arguably is a letter in German. That makes the key space of option b even bigger compared to the one of option c.*

The best we can do for option c if we restrict ourselves to the ascii alphabet is to use every printable character (excluding the blank) in our alphabet (note: different, more general use of the word "alphabet"). That's 94 characters, giving option c a key space of 94^7 ≈ 6.48*10^13 elements.

Since one of our assumptions to tackle the question is that the password is chosen randomly witch the respective restrictions and that rule is equal to choosing a password randomly from the respective key space, a password chosen using option b is arguably harder to guess since there are several orders of magnitude more options to try when cracking the password.

In fact, if we assume the costs of cracking a password via brute force to be approximately linear to the size of the key space, cracking a password chosen via option b is 52^10/(94^7) ≈ 2'229 times as hard as cracking one chosen via option c, clearly showing that the allegedly correct answer to this question is wrong.

* This is quite easy to prove mathematically but this StackExchange lacks LaTeX support and you probably will understand it better through a textual description anyways.

The only advantage option c as over option b is its bigger alphabet (again, more general use of the word "alphabet"). Option b, however, makes more than up for this by having choosing a longer password. If we add more and more characters (like ü, à, Ø, Æ, etc.) to it, we're making the alphabets more equal in size, causing the advantage of c over b to diminish, whereas the advantage of b over c is unaffected.