Answer
$-1.0\;\rm N\cdot m^2/C$
Work Step by Step
We know that the net electric flux through the closed surface is given by
$$\Phi=\dfrac{Q_{in}}{\epsilon_0}$$
And we are given that $Q_{in}=Nq_e$ where $N$ is the number of electrons and $q_e$ is the charge of one electron.
$$\Phi=\dfrac{Nq_e}{\epsilon_0}$$
Plug the known;
$$\Phi=\dfrac{(55.3\times 10^6)(-1.6\times 10^{-19})}{(8.85\times 10^{-12})}$$
$$\Phi=\color{red}{\bf -1.0}\;\rm N\cdot m^2/C$$