Answer
$q_{enc} = 7.29\times 10^{-11}~C$
Work Step by Step
$E = (3.00 y \hat{j})~N/C$
Note that that electric field is directed in the positive y direction when $0 \lt y$, and $E = 0$ when $y = 0$
Thus the electric field only passes through the righthand surface of the cube. The flux is zero everywhere else.
We can find the flux:
$\Phi = E~A~cos~\theta$
$\Phi = [(3.00)(1.40)~N/C]~(1.40~m)^2~cos~0^{\circ}$
$\Phi = 8.23~N~m^2/C$
We can find the enclosed charge:
$q_{enc} = \epsilon_0~\Phi$
$q_{enc} = (8.854\times 10^{-12}~F/m)(8.23~N~m^2/C)$
$q_{enc} = 7.29\times 10^{-11}~C$
