Physics or Maths?

Since the charge density is spherically symmetric, so is the electrostatic potential, and hence the electric field will be of the form $\mathbf{E} \; = \; F(r)\hat{\mathbf{r}}$ where $\hat{\mathbf{r}}$ is the radial unit vector. Thus, integrating over the surface of a sphere $S_R$ of radius $R$ centred at the origin, $4\pi R^2F(R) \; = \; \oint_{\partial S_R} \mathbf{E} \cdot d\mathbf{S} \; = \; 4\pi \int_{S_R} f\,dV \; = \; 4\pi \int_0^R \int_0^\pi \int_0^{2\pi} \frac{r^2\sin\theta}{1+r^2}\,d\phi\,d\theta\,dr \; = \; 16\pi^2(R - \tan^{-1}R)$ so that $F(R) \; = \; \frac{4\pi}{R^2}\big(R - \tan^{-1}R\big)$ The required answer to the question is $F(3) = \boxed{2.444793306}$ .