Answer
The net charge contained by the cube is $~~-1.70\times 10^{-9}~C$
Work Step by Step
Since the electric field is uniform in the x direction and the z direction, the net electric flux through the front and back faces is zero, and the net electric flux through the top and bottom faces is zero.
We can find the electric flux through the right face where $y = 4.00$:
$\Phi = (-4.00y^2)~\hat{j}\cdot A~\hat{j}$
$\Phi = (-64.0)~\hat{j}\cdot A~\hat{j}$
$\Phi = -64.0~A$
$\Phi = (-64.0~N/C)~(2.00~m)^2$
$\Phi = -256~N~m^2/C$
We can find the electric flux through the left face where $y = 2.00$:
$\Phi = (-4.00y^2)~\hat{j}\cdot (-A~\hat{j})$
$\Phi = (-16.0)~\hat{j}\cdot (-A~\hat{j})$
$\Phi = 16.0~A$
$\Phi = (16.0~N/C)~(2.00~m)^2$
$\Phi = 64.0~N~m^2/C$
The net flux through the cube is $-192~N~m^2/C$
We can find the net charge contained by the cube:
$q = \epsilon_0~\Phi$
$q = (8.854\times 10^{-12}~F/m)(-192~N~m^2/C)$
$q = -1.70\times 10^{-9}~C$
The net charge contained by the cube is $~~-1.70\times 10^{-9}~C$
