Answer
$r = 2~R$
Work Step by Step
We can find the magnitude of the electric field a distance $R$ from the center:
$\epsilon_0~\Phi = q$
$(\epsilon_0)~(E)(4\pi~R^2) = q$
$E= \frac{q}{4\pi~\epsilon_0~R^2}$
Suppose $r \gt R$
Note that the enclosed charge is $q$
We can find an expression for the magnitude of the electric field a distance $r$ from the center where $r \gt R$:
$\epsilon_0~\Phi = q$
$(\epsilon_0)~(E_r)(4\pi~r^2) = q$
$E_r= \frac{q}{4\pi~\epsilon_0~r^2}$
Suppose $E_r = 0.25~E$
We can find an expression for $r$:
$E_r = 0.25~E$
$\frac{q}{4\pi~\epsilon_0~r^2} = 0.25~\frac{q}{4\pi~\epsilon_0~R^2}$
$\frac{1}{r^2} = \frac{0.25}{R^2}$
$r^2 = 4~R^2$
$r = 2~R$
