Answer
$E = 0.112~N/C$
Work Step by Step
We can find the magnitude of the electric field at $r = a$:
$\phi = \frac{q_{enc}}{\epsilon_0}$
$E~(4\pi~r^2) = \frac{q_{enc}}{\epsilon_0}$
$E = \frac{q_{enc}}{4\pi~a^2~\epsilon_0}$
$E = \frac{5.00\times 10^{-15}~C}{(4\pi)~(0.0200~m)^2~(8.854\times 10^{-12}~F/m)}$
$E = 0.112~N/C$
