$\large\frac{r_{O_2}}{r_x} = \sqrt{\large\frac{M_x}{M_{O_2}}}$
Volume are in the same ratio , the no. of moles:
$=\large\frac{n/18}{n/45} = \sqrt{\large\frac{M_x}{32}}$
$\large\frac{45}{18} = \sqrt{\large\frac{M_x}{32}}$
$\therefore M_x = \large\frac{45^2}{18^2} \times 32$
= 200
Hence answer is (A)