Escape velocity of object from planet is given by

$$\begin{aligned} & \left(v_e\right)_p=\sqrt{\frac{2 G M_p}{R_p}} \\ & \therefore \quad\left(v_e\right)_p \propto \sqrt{\frac{M_p}{R_p}} \end{aligned}$$

Now, $M_p=9 M_e$ and $R_\rho=16 R_e$ (given)

$$\begin{aligned} & \frac{\left(v_e\right)_p}{\left(v_e\right)_e}=\sqrt{\frac{M_p}{R_p} \times \frac{R_e}{M_e}}=\sqrt{\frac{9 M_e}{16 R_e} \times \frac{R_e}{M_e}}=\sqrt{\frac{9}{16}}=\frac{3}{4} \\ & \Rightarrow \quad\left(v_e\right)_p=\frac{3}{4} v \end{aligned}$$