(12!)^2 is wrong for a couple of reasons.
1. The question states that the "insect" has a shoe and sock for each of its 12 legs. That is, he has a shoe and sock already assigned to each leg. This problem is not about how we can distribute 12 socks to each of his 12 legs and then 12 shoes to each of his 12 legs. It's about how many different orders he can put on his 12 socks and 12 shoes that he has already picked out for his 12 legs. Thus, there is no danger of him putting two shoes on the same foot because each shoe (and sock) is already assigned to a specific leg. To illustrate, let's look at a simpler case of something with three legs and a sock and shoe for each leg. Let A,B, and C represent the socks for legs 1, 2 and 3, respectively. Let 1,2,3 represent the shoes for legs 1, 2, and 3 respectively. So, one possibility is the sequence ABC123 which would represent him putting all of his socks on first in order and then all of his shoes on in order. This sequence can be arranged 6! ways. But in half of these, he puts his shoe on over his sock on his first foot. In another half he puts his shoe on over his sock on his second foot, and so on. So we have to divide by 2^3 to divide out these sequences that violate the restrictions
2. So maybe you're thinking, "hey, this question is ambiguous. it's not clear that each sock and shoe are preassigned to specific legs." Even if I grant you that, 12!^2 still doesn't make any sense (and this is the bigger point) because it doesn't take into account the order in which he puts them on. It conspicuously ignores the fact that the insect doesn't have to put on ALL of his socks before putting on his shoes; he only has to make sure that he put a sock on any given leg first before he puts a shoe on that leg. There's no reason why he can't, say, put on his sock and shoe for his first leg before putting any other socks or shoes on. For example, let's say that you have a red sock and a blue sock, and two different shoes that can (remarkably) go on either of your feet. We can surely agree that there are 2*2=4 ways to assign a sock and a shoe to each foot. But once you've assigned the socks and shoes, there are still many different orders in which you could put them on your feet. Specifically, there are six possibilities:
1. Right sock, left sock, right shoe, left shoe
2. Right sock, left sock, left shoe, right shoe
3. Right sock, right shoe, left sock, left shoe
4. Left sock, right sock, right shoe, left shoe
5. Left sock, right sock, left shoe, right shoe
6. Left sock, left shoe, right sock, right shoe
So there would be 4 (or 2!2!) ways to assign the socks and shoes and then 6 (or 4!/2^2) orders in which you could put any given assignment of socks and shoes on. This question asks about the order, and never mentions having to first select a sock and shoe for each leg, so 24!/2^12 is the right answer.
Now, the problem COULD be modified so that 12!^2 makes sense. For example, if all of the socks and shoe are identical (or, if they're not identical but are pre-assigned to specific legs) and you add the restriction that the insect has to COMPLETELY FINISH putting his socks on before putting his shoes on, then when he chooses the order for his socks, he would have 12 choices for the first leg, 11 for the second, 10 for third, and so on or 12! orders for putting his socks on. And then when he was done, he'd put his 12 shoes on by the same process and have 12! choices for the order in which he did. But again, the key is that the problem, as originally stated, never says the insect has to put ALL of his socks on first, just that a sock has to go on any given leg before the shoe goes on that leg.
Anyway, C is the correct answer.