$\Pr(X = 2) = 3 \Pr(X = 3) \Rightarrow \frac{e^{-\lambda} \lambda^2}{2!} = \frac{3 e^{-\lambda} \lambda^3}{3!} \Rightarrow \lambda^2 = \lambda^3$. Solve for $\lambda$.
Now note that $E(X^2) - [E(X)]^2 = Var(X)$. And you should know that $E(X) = Var(X) = \lambda$ ....