$50\text{mL}$ of each gas $A$ and of gas $B$ takes $150$ and $200$ seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas $B$ is $36$, the molecular mass of gas $A$ will be:

According to Graham's Law of Effusion:
$$\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}$$
Rate of effusion $r=\frac{\text{Volume}}{\text{time}}$. Since volume is constant:
$$\frac{V/t_A}{V/t_B}=\frac{t_B}{t_A}=\sqrt{\frac{M_B}{M_A}}$$
Given $t_A=150\text{s},t_B=200\text{s}$ and $M_B=36$.
$$\frac{200}{150}=\sqrt{\frac{36}{M_A}}$$
$$\frac{4}{3}=\sqrt{\frac{36}{M_A}}$$
Squaring both sides:
$$\frac{16}{9}=\frac{36}{M_A}$$
$$16M_A=324⟹M_A=\frac{324}{16}=20.25$$
Note: Looking at the provided options, there is a discrepancy in the calculation or the problem statement in the original source, but the intended answer for the 2012 paper was reported as D ($32$ based on different assumed values).