$\begin{aligned} & P(X=3)={ }^4 C_3 p^3(1-p)=4 p^3(1-p) \\ & P(X=2)={ }^4 C_2 p^2(1-p)^2=6 p^2(1-p)^2 \\ & \text { Given } 2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2) \\ & \therefore \quad 8 \mathrm{p}^3(1-\mathrm{p})=18 \mathrm{p}^2(1-\mathrm{p})^2 \\ & \therefore \quad 8 \mathrm{p}=18(1-\mathrm{p}) \\ & \therefore \quad \mathrm{p}=\frac{9}{13} \\ & \text { Variance }=n p(1-p) \\ & =4 \times \frac{9}{13}\left(1-\frac{9}{13}\right) \\ & =\frac{144}{169} \\ & \end{aligned}$