Since the gas took $\large\frac{50}{9.62}$ times longer than Hydrogen its rate of diffusion would be $\large\frac{9.62}{50}$ times that of Hydrogen.

$\Rightarrow \large\frac{9.62}{50} = \large\sqrt \frac{m_{H_2}}{m_{gas}}$

Given $m_{H_2} = 2 \rightarrow m_{gas} = \large(\frac{50}{9.62})$$^2 \times 2 = 54$

Among all the options presented to us above, the one option with a molar mass of $54$ is (D) $C_2H_2N_2$.

Note: We need not use the data of composition of gas.