Electromagnetism - Gauss' Law

A sphere of radius $R$ carries charge density proportional to the square of the distance from the center: $\rho=Ar^2$, where $A$ is a positive constant. At a distance of $R/2$ from the center, the magnitude of the electric field is:
A. $A/4\pi\varepsilon_0$
B. $AR^3/40\varepsilon_0$
C. $AR^3/24\varepsilon_0$
D. $AR^3/5\varepsilon_0$
E. $AR^3/3\varepsilon_0$

(GR9677 #61)

Solution:

Gauss’ Law: $\oint \bar{E} \cdot \hat{n} dA=\frac{q_{\textup{enc}}}{\varepsilon_0}$
$\rightarrow E\cdot 4\pi \bigg(\frac{R}{2}\bigg)^2=\frac{q_{\textup{enc}}}{\varepsilon_0}$
$\rightarrow E=\frac{q_{\textup{enc}}}{\pi R^2 \varepsilon_0}$
The net charge within the Gussian surface with $\rho$ as a function of $r$ (non-uniformly charged sphere):
$dq_{\textup{enc}}=\rho dV=Ar^2 d\bigg(\frac{4}{3} \pi r^3 \bigg)$
$dq_{\textup{enc}}=Ar^2 \frac{4}{3} \pi 3r^2 dr=A4 \pi r^4 dr$
$q_{\textup{enc}}=\int_0^{R/2} A4\pi r^4 dr=A4 \pi \frac{1}{5} \bigg(\frac{R}{2}\bigg)^5=\frac{A\pi R^5}{40}$
$\rightarrow E=\frac{1}{\pi R^2 \varepsilon_0}\cdot \frac{A \pi R^5}{40}=\frac{AR^3}{40\varepsilon_0}$
Answer: B