I'll try to solve it, but am unsure about my understanding. I think it's 450*y=n^3
If so, what I can do is prime factor 450y:
2*3*3*5*5*y=n*n*n
It means, that all of the prime factors need to appear 3 times, so y needs to have at least 2*2*3*5.

If we divide it by (3*22*5) we are dividing it by (2*3*5*11). In order for it to be an integer, y needs to contain at least 1 11 in it, and in order for it to meet the condition, it need 3 11s. It is possible that y has 3 11s in it, but we don't know it, so we cannot say for sure that it is an integer.

If we divide it by (32*2*5) we actually divide by (2*2*2*2*2*2*5). We know that y has two 2s, but don't know whether it has the additional 4(four) 2s to be an integer and additional 6 2s to meet the equation.

In the third option, we divide by (2*2*2*3*13), and again we don't know whether we have additional 2s or 13s in y.

I would go for A(none), since although it is possible for y to contain additional integers, we know nothing about it and thus cannot be sure of it. I am, however, unsure about it.