Answer
$\Phi = -48~N~m^2/C$
Work Step by Step
$E = 4.0 \hat{i}-3.0(y^2+2.0)~\hat{j}$
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 found in part (c) that the electric flux through the left face is $~~24~N~m^2/C$
We can find the electric flux through the right face where $y=2.0$:
$\Phi_r = -3.0(y^2+2.0)~\hat{j}~\cdot A~\hat{j}$
$\Phi_r = -3.0(2.0^2+2.0)~\hat{j}~\cdot A~\hat{j}$
$\Phi_r = -(18)~(2.0~m)^2$
$\Phi_r = -72~N~m^2/C$
We can find the net electric flux:
$\Phi = (24~N~m^2/C)+(-72~N~m^2/C) = -48~N~m^2/C$
