Answer
$113\;\rm N\cdot m^2/C$
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 the cylinder includes only the 1 nC charge inside it.
So that the net electric flux through the cylinder is given by
$$\Phi=\dfrac{ 1\times 10^{-9}}{8.85\times 10^{-12}}$$
$$\Phi=\color{red}{\bf 113}\;\rm N\cdot m^2/C$$
