The answer is D: 75.

This is a question relating to prime factorability (is that a word?).

Basically, for any number, X, to be a factor of another number, Y, the prime factors of X must be present in the prime factors of Y. For example, if you reduce the number 50 = (5^2)*(2^1), it is a factor of any number which has these prime factors when reduced.

So, for 5^2 and 3^3 to be factors of this number, the number, when reduced to prime factors, must contain 5^2 and 3^3.

n*2^5*6^2*7^3

= n*2^5*(3^2*2^2)*7^3

= n*2^7*3^2*7^3

As seen by the above equation, we need at least one more factor of 3, as well as two more factors of 5. As a result:

n = 5^2 * 3

= 75