Answer
$N = 1.35 \times 10^{10} \mathrm{electron}$
Work Step by Step
First, we will find the total charge $q$ where the electric field is related to the charge $q$ by
\begin{gather}
E=\frac{1}{4 \pi \epsilon_{\odot}} \frac{q}{r^{2}}\\
q = 4 \pi \epsilon_{\circ} r^{2} E \\
q =\frac{(1150 \mathrm{N} / \mathrm{C})}{9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}\left(13 \times 10^{-2} \mathrm{m}\right)^{2}\\
q =2.16 \times 10^{-9} \mathrm{C}
\end{gather}
Now, we divide this charge by the charge of electrons to get the number of the excess electrons
\begin{aligned}
N &=\frac{q}{e} \\
&=\frac{\left(2.16 \times 10^{-9} \mathrm{C}\right)}{1.6 \times 10^{-19} \mathrm{C} / \mathrm{electron}} \\
&=1.35 \times 10^{10} \mathrm{electron}
\end{aligned}