Escape velocity, expressed as

$$v_e = \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} = \sqrt{\frac{2 \mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^3 \rho}{\mathrm{R}}}$$

can be simplified to

$$\sqrt{\frac{8 \pi \mathrm{G} \rho}{3}} \mathrm{R}.$$

The surface area, $A_s$, is given by

$$A_s = 4 \pi R^2.$$

Considering planet Q's surface area is 4 times that of planet P:

$$\mathrm{A}_{\mathrm{Q}} = 4 \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{Q}}^2$$

$$\mathrm{~A}_{\mathrm{P}} = \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{P}}^2$$

therefore,

$$4 \mathrm{R}_{\mathrm{P}}^2 = \mathrm{R}_{\mathrm{Q}}^2.$$

Solving for $R_Q$ :

$$\Rightarrow \quad R_{\mathrm{Q}} = 2 \mathrm{R}_{\mathrm{P}} \quad \text{........(i)}.$$

For planet R, with mass $\mathrm{M}_{\mathrm{R}} = \mathrm{M}_{\mathrm{P}} + \mathrm{M}_{\mathrm{Q}}$:

$$\rho \cdot \frac{4}{3} \pi \mathrm{R}_{\mathrm{R}}^3 = \rho \cdot \frac{4}{3} \pi (\mathrm{R}_{\mathrm{P}}^3 + \mathrm{R}_{\mathrm{Q}}^3)$$

Since

$$\mathrm{R}_{\mathrm{Q}} = 2\mathrm{R}_{\mathrm{P}},$$

we have :

$$\mathrm{R}_{\mathrm{R}}^3 = \mathrm{R}_{\mathrm{P}}^3 + 8 \mathrm{R}_{\mathrm{P}}^3 = 9 \mathrm{R}_{\mathrm{P}}^3.$$

Thus,

$$\mathrm{R}_{\mathrm{R}} = (9)^{1/3} \mathrm{R}_{\mathrm{P}} \quad \text{........(ii)}.$$

From (i) and (ii), it follows that :

$$ R_R > R_Q > R_P.$$

Since

$$V_e \propto \mathrm{R},$$

it implies that :

$$\mathrm{V}_{\mathrm{R}} > \mathrm{V}_{\mathrm{Q}} > \mathrm{V}_{\mathrm{P}}.$$

Also,

$$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{P}}} = \frac{\mathrm{R}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{P}}} = 9^{1/3},$$

and

$$\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{Q}}} = \frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}} = \frac{1}{2}.$$