[$$:]\oint \vec{E} \cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}[/:$$] This is Gauss' Law. And you are given the charge density as: [$$:]\rho = \rho_0 \frac{1-r^2}{a^2}[/:$$] Remember that charge density is defined as: [$$:]\rho = \frac{Q}{V}=\frac{dq}{dV}[/:$$] Hence, we have: [$$:]dq=\rho \; dV[/:$$] Well, we can find [$:]dV[/:$]. We are talking about a sphere here, so it will be: [$$:]dV=4\pi r^2 dr[/:$$] That will be the micro volume. You can draw a diagram to get this, but this is very simply the derivative of the volume of a sphere, which is [$:]\frac{4}{3}\pi r^3[/:$]. Plug everything back in to [$:]dq[/:$], including the density: [$$:]dq = \rho_0 \frac{1-r^2}{a^2} 4\pi r^2 dr=\frac{\rho_0 4\pi}{a^2}(r^2-r^4) dr[/:$$] From this, you can get [$:]q_{enc}[/:$]. [$:]q_{enc}[/:$] is just the integral. Let's start with [$:]0 \le r < a[/:$]. The charge will be: [$$:]q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^r (r^2-r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}-\frac{r^5}{5}\right)[/:$$] The thing about Gauss's law is that, in most cases, the electric field won't be dependent on the area, hence: [$$:]\oint \vec{E} \cdot d\vec{A}=\vec{E} \cdot \oint d\vec{A}=\vec{E}\cdot\vec{A}[/:$$] The area is very simple. It's just [$:]4\pi r^2[/:$] because it's a sphere. Therefore: [$$:]E(4 \pi r^2)=\frac{\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}-\frac{r^5}{5}\right)}{\epsilon_0}[/:$$] Simplifying gives you: [$$:]\therefore E=\frac{\rho_0}{a^2\epsilon_0} \left(\frac{r}{3}-\frac{r^3}{5}\right)[/:$$] This is for [$:]0 \le r < a[/:$]. Similar thing for when it's greater than [$:]a[/:$], except you will need the sphere's full charge: [$$:]q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^a (r^2-r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{a^3}{3}-\frac{a^5}{5}\right)=\rho_0 4\pi \left(\frac{a}{3}-\frac{a^3}{5}\right)[/:$$] We only integrate to [$:]a[/:$] because afterwards there's no charge. That's the sphere's total enclosed charge. *However*, the Gaussian surface's area is still [$:]4 \pi r^2[/:$], of course. Hence we have: [$$:]E(4 \pi r^2)=\frac{\rho_0 4\pi \left(\frac{a}{3}-\frac{a^3}{5}\right)}{\epsilon_0}[/:$$] Simplified to: [$$:]\therefore E=\frac{\rho_0}{\epsilon_0 r^2} \left(\frac{a}{3}-\frac{a^3}{5}\right)[/:$$]