Answer
$E = 25.3~N/C$
Work Step by Step
We can find the charge inside a sphere of radius $r = 3.00~cm$:
$q = \frac{(3.00~cm)^3}{(4.00~cm)^3}(6.00~pC)$
$q = 2.53125\times 10^{-12}~C$
We can find the electric field at a $r = 3.00~cm$:
$\epsilon_0~\Phi = q$
$(\epsilon_0)~(E)~(4\pi~r^2) = q$
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$E = \frac{2.53125\times 10^{-12}~C}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0300~m)^2}$
$E = 25.3~N/C$
