Answer
We show that the flux does not depend on the radius of the sphere, that is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = 4\pi $
Work Step by Step
Let the sphere ${x^2} + {y^2} + {z^2} = {R^2}$ be parametrized by
$G\left( {\theta ,\phi } \right) = \left( {R\cos \theta \sin \phi ,R\sin \theta \sin \phi ,R\cos \phi } \right)$
By Eq. (2) in Section 17.4, the outward-pointing normal vector is ${\bf{N}}\left( {\theta ,\phi } \right) = {R^2}\sin \phi {{\bf{e}}_r}$.
We evaluate the flux of ${\bf{F}} = \frac{{{{\bf{e}}_r}}}{{{r^2}}}$ through the sphere by Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $
Since the radius of the sphere is $R$, so ${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right) = \frac{{{{\bf{e}}_r}}}{{{R^2}}}$. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {\frac{{{{\bf{e}}_r}}}{{{R^2}}}} \right)\cdot\left( {{R^2}\sin \phi {{\bf{e}}_r}} \right){\rm{d}}\theta {\rm{d}}\phi $
Notice that the term ${R^2}$ in the numerator and the denominator cancel out. So,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\theta {\rm{d}}\phi $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi = 2\pi \left( { - \cos \phi } \right)|_0^\pi = 4\pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = 4\pi $.
We see that the flux does not depend on the radius of the sphere.