Answer
The flux of ${\bf{E}}$ through the disk $D$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = 6\pi k\left( {\dfrac{1}{3} - \dfrac{1}{{\sqrt {13} }}} \right)$
Work Step by Step
Using the polar coordinates: $x = \rho \cos \theta $ and $y = \rho \sin \theta $, the disk $D$ can be parametrized by $G\left( {\rho ,\theta } \right) = \left( {\rho \cos \theta ,\rho \sin \theta ,3} \right)$, where $0 \le \rho \le 2$, $0 \le \theta \le 2\pi $.
So,
${{\bf{T}}_\rho } = \left( {\cos \theta ,\sin \theta ,0} \right)$
${{\bf{T}}_\theta } = \left( { - \rho \sin \theta ,\rho \cos \theta ,0} \right)$
${\bf{N}}\left( {\rho ,\theta } \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\cos \theta }&{\sin \theta }&0\\
{ - \rho \sin \theta }&{\rho \cos \theta }&0
\end{array}} \right| = \rho {\bf{k}}$
We evaluate the flux of ${\bf{E}}$ through the disk $D$ using Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\left( {G\left( {\rho ,\theta } \right)} \right)\cdot{\bf{N}}\left( {\rho ,\theta } \right){\rm{d}}\rho {\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} \left( {k\dfrac{{{{\bf{e}}_r}}}{{{r^2}}}} \right)\cdot\left( {\rho {\bf{k}}} \right){\rm{d}}\rho {\rm{d}}\theta $
Recall from Exercise 19, that ${{\bf{e}}_r} = \left( {\dfrac{x}{r},\dfrac{y}{r},\dfrac{z}{r}} \right)$. So, ${{\bf{e}}_r}\cdot{\bf{k}} = \left( {\dfrac{x}{r},\dfrac{y}{r},\dfrac{z}{r}} \right)\cdot\left( {0,0,1} \right) = \dfrac{z}{r}$.
Since the disk is on $z=3$, we get ${{\bf{e}}_r}\cdot{\bf{k}} = \dfrac{3}{r}$. So, the integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = 3k\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} \dfrac{\rho }{{{r^3}}}{\rm{d}}\rho {\rm{d}}\theta $
But ${r^2} = {x^2} + {y^2} + {z^2} = {\rho ^2} + {z^2}$ and $z=3$. So, ${r^3} = {\left( {{\rho ^2} + 9} \right)^{3/2}}$. Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = 3k\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\rho = 0}^2 \dfrac{\rho }{{{{\left( {{\rho ^2} + 9} \right)}^{3/2}}}}{\rm{d}}\rho {\rm{d}}\theta $
Write $t = {\rho ^2} + 9$. So, $dt = 2\rho d\rho $. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = \dfrac{3}{2}k\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{t = 9}^{13} \dfrac{1}{{{t^{3/2}}}}{\rm{d}}t$
$ = 3\pi k\left( { - 2{t^{ - 1/2}}} \right)|_9^{13} = - 6\pi k\left( {\dfrac{1}{{\sqrt {13} }} - \dfrac{1}{3}} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_D^{} {\bf{E}}\cdot{\rm{d}}{\bf{S}} = 6\pi k\left( {\dfrac{1}{3} - \dfrac{1}{{\sqrt {13} }}} \right)$.
