${{\rho }_{A}}=1.5{{\rho }_{B}}$
${{\rho }_{A}}=2{{\rho }_{B}}$
According to ideal gas equation, we have Pressure,$p=\frac{\rho RT}{M}$
, where M is molecular weight of ideal gas.
Such that, $\frac{p}{\rho }=\frac{RT}{M}\Rightarrow M=\frac{\rho RT}{p}$
where, R and T are constant.
So, $M\propto \frac{\rho }{p}$
$\Rightarrow \,\,\frac{{{M}_{A}}}{{{M}_{B}}}=\frac{{{\rho }_{A}}}{{{\rho }_{B}}}\times \frac{{{p}_{B}}}{{{p}_{A}}}=1.5\times \frac{1}{2}=0.75=\frac{3}{4}$