Answer
$r = \sqrt{2}~R$
Work Step by Step
Let $q$ be the magnitude of the total negative charge in the sphere.
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 $F_r = 0.50~F_R$:
$F_r = \frac{q_p~q}{4\pi~\epsilon_0~r^2} = 0.50~F_R$
$\frac{q_p~q}{4\pi~\epsilon_0~r^2} = 0.50~\frac{q_p~q}{4\pi~\epsilon_0~R^2}$
$\frac{1}{r^2} = 0.50~\frac{1}{R^2}$
$r^2 = 2~R^2$
$r = \sqrt{2}~R$
