Answer
$2.88\times 10^{4}\ N/C$
Work Step by Step
Gauss law relates the net flux $\phi$ of an electric field through a closed surface (a Gaussian surface) to net charge $q_{enc}$ that is enclosed by that surface. it tells that
$\epsilon_{o} \phi = q_{enc}$
$\epsilon_{o}\oint E.dA = q_{enc}$
$\epsilon_{o} E(4\pi r^2) = q$
$E = \frac{1}{4\pi\epsilon_{o}}\frac{q}{r^2}$
The electric field E at R=25 cm is thus:
$E = 9\times 10^{9}\frac{(2\times 10^{-7})}{(0.25^2)}$
$E = 2.88\times 10^{4} N/C$
