Answer
The two masses go up to a height of $\frac{R}{4}$.
Work Step by Step
We can find the falling block's speed just before the collision.
$K = PE$
$\frac{1}{2}mv_1^2 = mgR$
$v_1 = \sqrt{2gR}$
We can use conservation of momentum to find the speed just after the collision.
$(2m)v_2 = mv_1$
$v_2 = \frac{v_1}{2}$
$v_2 = \frac{\sqrt{2gR}}{2}$
$v_2 = \sqrt{\frac{gR}{2}}$
The potential energy at the highest point will be equal to the kinetic energy at the bottom.
$PE = K$
$(2m)gh = \frac{1}{2}(2m)v_2^2$
$h = \frac{v_2^2}{2g}$
$h = \frac{(\sqrt{\frac{gR}{2}})^2}{2g}$
$h = \frac{(\frac{gR}{2})}{2g}$
$h = \frac{R}{4}$
The two masses go up to a height of $\frac{R}{4}$.
