Answer
$ \Phi_E = -0.98 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$. Inward
Work Step by Step
Using Gauss's law we could get the electric flux for the enclosed surface by
\begin{align}
\Phi_{E}&=\frac{Q}{\epsilon_{0}}\\
&=\frac{-8.65 \times 10^{-12} \mathrm{C}}{8.854 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)}\\
&=\boxed{ -0.98 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}}
\end{align}
The electric flux is inward as the charge is negative,
