Answer
$q = 3.74 \times 10^{-10} \mathrm{C}$
Work Step by Step
First, we find the electric flux then the charge. The electric flux equals the area time the electric field through this area where the area $A$ equals $ 2 \pi r l $
$\begin{aligned}
\Phi_{E} &= EA \\
&= 2 \pi r E l \\
&=2 \pi(0.4 \mathrm{m})(840 \mathrm{N} / \mathrm{C})(0.020 \mathrm{m}) \\ &=42.2 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}
\end{aligned}$
Now we could get the charge $q$ by
\begin{aligned}
q &=\epsilon_{0} \Phi_{E} \\ &=\left(8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)\left(42.2 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\right) \\ &=\boxed{3.74 \times 10^{-10} \mathrm{C}}
\end{aligned}
