According to the Ideal Gas Equation, $PV = nRT$. The number of moles $n$ can be written as $\frac{\text{mass } (m)}{\text{Molar Mass } (M)}$. Substituting this into the gas equation: $PV = \frac{m}{M}RT$. Rearranging to introduce density ($\rho = \frac{m}{V}$): $P = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M}$. Solving for Molar Mass ($M$): $M = \frac{\rho RT}{P}$.

Given conditions:

1. Temperature is constant ($T_A = T_B$).
2. Pressure of A is twice that of B: $P_A = 2P_B \Rightarrow \frac{P_B}{P_A} = \frac{1}{2}$.
3. Density of A is 1.5 times that of B: $\rho_A = 1.5 \rho_B \Rightarrow \frac{\rho_A}{\rho_B} = 1.5 = \frac{3}{2}$.

Calculation: Ratio of molecular weights $\frac{M_A}{M_B} = \frac{\rho_A RT / P_A}{\rho_B RT / P_B} = \left( \frac{\rho_A}{\rho_B} \right) \times \left( \frac{P_B}{P_A} \right)$. $\frac{M_A}{M_B} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$.