Answer
$\Phi = 8.23~N~m^2/C$
Work Step by Step
$E = [-4.00 \hat{i}+(6.00+3.00y)~\hat{j}]~N/C$
Note that the electric flux through the top face and the bottom face is zero.
Since the x component of the electric field is constant, the electric flux through the front face and the back face cancel out for a net electric flux of zero.
We can find the electric flux through the left face where $y=0$:
$\Phi_l = (6.00+3.00y)~\hat{j}~\cdot -A~\hat{j}$
$\Phi_l = -6.00~A$
$\Phi_l = -(6.00)~(1.40~m)^2$
$\Phi_l = -11.76~N~m^2/C$
We can find the electric flux through the right face where $y=1.40~m$:
$\Phi_r = (6.00+3.00y)~\hat{j}~\cdot A~\hat{j}$
$\Phi_r = [6.00+(3.00)(1.40)]~A$
$\Phi_r = (10.20)~(1.40~m)^2$
$\Phi_r = 19.992~N~m^2/C$
We can find the net electric flux:
$\Phi = (19.992~N~m^2/C)+(-11.76~N~m^2/C)$
$\Phi = 8.23~N~m^2/C$
