Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_P$ and $M_Q$, and surface areas $A$ and $4A$, respectively. A spherical planet $R$ also has uniform density $\rho$ and its mass is $\left(M_P+M_Q\right)$. The escape velocities from the planets$P,Q$and $R$, are $v_P,v_Q$ and $v_R$. respectively. Then,

Surface area of $Q$ is four times. Therefore, radius of $Q$ is two times. Volume is eight times. Therefore, mass of $Q$ is also eight times.

So, let $M_P=M$ and $R_P=r$

Then, $M_Q=8M$ and $R_Q=2r$

Now, mass of $R$ is $\left(M_P+M_Q\right)$ or $9M$. Therefore, radius of $R$ is $(9{)}^{1/3}r$. Now, escape velocity from the surface of a planet is given by

       $v=\sqrt{\frac{2GM}{r}} (r=$ radius of that planet $)$

$\therefore$$v_p=\sqrt{\frac{2GM}{r}}$

    $v_Q=\sqrt{\frac{2G(8.f)}{\left(2r\right)}}$

    $v_R=\sqrt{\frac{2G\left(9M\right)}{(9{)}^{1/3}r}}$

From here, we can see that

$\frac{v_P}{v_Q}=\frac{1}{2}$ and $v_R>v_Q>v_P$