If $r, s$ and $t$ are prime numbers and $p, q$ are positive integers such that the LCM of $p,q$ is $\displaystyle r^{2}t^{4}s^{2}$ then the number of ordered pair $(p, q)$ is 

If 2 numbers A and B have LCM as L, where L = ${ a }^{ x }\times{ b }^{ y }\times{ c }^{ z }$ ., then, number of ordered pairs of numbers having the above LCM will be: $(2x + 1) \times (2y + 1) \times (2z + 1)$

Let us consider r: we need the power r in either p or q to be at least $2$.

If the power of r in p is 2, then in q it should be 0 or 1 or 2 $\rightarrow3$ cases

If the power of r in q is 2, then in q it should be 0 or 1 or 2 $\rightarrow3$ cases

But (2, 2) has been counted twice. Thus, there are $3 + 3 - 1 = 5$ cases

Total of five cases for the exponents of r such that we have the given LCM which is $2\times 2 + 1$

Similarly, the others follow.

Here, $x = 2, y = 4, z = 2.$

Therefore, the answer = $(2\times2+1)\cdot (2\times4+1)\cdot (2\times2+1)=225$

Hence, (C) is correct.