Answer
$r = 0.25~R$
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 $r \lt R$
We can find an expression for the enclosed charge:
$q_{enc} = \frac{r^3}{R^3}~q$
We can find an expression for the magnitude of the electric field a distance $r$ from the center where $r \lt R$:
$\epsilon_0~\Phi = q_{enc}$
$(\epsilon_0)~(E_r)(4\pi~r^2) = \frac{r^3}{R^3}~q$
$(\epsilon_0)~(E_r)(4\pi) = \frac{r}{R^3}~q$
$E_r = \frac{q~r}{4\pi~\epsilon_0~R^3}$
Suppose $E_r = 0.25~E$
We can find an expression for $r$:
$E_r = 0.25~E$
$\frac{q~r}{4\pi~\epsilon_0~R^3} = 0.25~\frac{q}{4\pi~\epsilon_0~R^2}$
$\frac{r}{R} = 0.25$
$r = 0.25~R$
