For $\mathrm{A}, \frac{2}{M w_{A}} \propto 1 \mathrm{atm}$ $...(i)$

Pressure of $B=1.5-1=0.5$ atm

For $\mathrm{B}, \frac{3}{M w_{B}} \propto 0.5 \mathrm{atm}$  $...(ii)$

Dividing $(ii)$ by $(i),$ we get,

$\frac{3}{M w_{B}} \times \frac{M w_{A}}{2}=\frac{0.5}{1}$

$\frac{M w_{A}}{M w_{B}}=0.5 \times \frac{2}{3}=\frac{1}{3}$

$\therefore M w_{A}: M w_{B}=1: 3$

Therefore, the option is $A$