Answer
$\Phi = 0.50~N~m^2/C$
Work Step by Step
We can find the electric field at $x = 2.0~m$:
$E = (x+2)~\hat{i}~N/C$
$E = (2+2)~\hat{i}~N/C$
$E = (4.0~\hat{i})~N/C$
We can find the flux through the end of the cylinder at $x = 2.0~m$:
$\Phi = E~\cdot A$
$\Phi = E~\pi~r^2$
$\Phi = (4.0~N/C)~(\pi)~(0.20~m)^2$
$\Phi = 0.50~N~m^2/C$
