Answer
$r = 0.50~R$
Work Step by Step
Let $q$ be the magnitude of the total negative charge in the sphere.
We can find the magnitude of the negative charge within a sphere of radius $r$ where $r \lt R$:
$q_r = \frac{r^3}{R^3}~q$
We can write an expression for $F_R$:
$F_R = \frac{q_p~q}{4\pi~\epsilon_0~R^2}$
We can find $r$ when $F_r = 0.50~F_R$:
$F_r = \frac{q_p~q_r}{4\pi~\epsilon_0~r^2} = 0.50~F_R$
$\frac{q_p~q~r^3}{4\pi~\epsilon_0~r^2~R^3} = 0.50~\frac{q_p~q}{4\pi~\epsilon_0~R^2}$
$\frac{r}{R} = 0.50$
$r = 0.50~R$
