Answer
$E = 0.0562~N/C$
Work Step by Step
We can find the enclosed charge in a sphere of radius $r$ when $r \leq a$:
$q_{enc} = \frac{r^3}{a^3}~q_1$
We can find an expression for the electric field at radius $r$ where $r \leq a$:
$\epsilon_0~\Phi = q_{enc}$
$(\epsilon_0)(E)(4\pi~r^2) = \frac{r^3~q_1}{a^3}$
$E = \frac{r~q_1}{4\pi ~\epsilon_0 ~a^3}$
We can find the electric field when $r = \frac{a}{2.00}$:
$E = \frac{r~q_1}{4\pi ~\epsilon_0 ~a^3}$
$E = \frac{(a/2.00)~q_1}{4\pi ~\epsilon_0 ~a^3}$
$E = \frac{q_1}{8\pi ~\epsilon_0 ~a^2}$
$E = \frac{5.00\times 10^{-15}~C}{(8\pi) (8.854\times 10^{-12}~F/m) ~(0.0200~m)^2}$
$E = 0.0562~N/C$
