Firstly, let me just remark that problems by McGraw-Hill are generally terrible. So don't take them as a very accurate gauge of how the GRE works.

A divisor is a number that can divide the number in question with no remainder. You should recognize 121 as 11^2, so its divisors are 1, 11, and 121. Don't count 11 twice! The trick in this question, and it's a pretty mild one, is that 24 is much smaller than 121, but has lots more divisors. It's got 1, 2, 3, 4... At this point you know the answer is B so you could stop but just for completion's sake, it's also got 6, 8, 12, and 24.

These were (overly) simple examples but a good method, by the way, of making sure you've got every divisor of a number is to make two columns showing what two numbers could multiply to make it. Always start at 1. That'll make sure you don't forget 1 and the number itself. So by this method the factors of 24 would be:

1 24
2 12
3 8
4 6

When the two numbers are the same, or cross over each other, you're done. Let's do 64:

1 64
2 32
4 16
8 8

So there are 7 total for 64 because 8 shouldn't count twice!