
Given here, electric field, E=E0xi^
Here, the flux passes mainly through surface areas, ABCDandEFGH. As the surfaces AEFBandCGHD are parallel to the electric field, so flux for these surfaces are zero. Again in EFGH, a=0, thus, electric field is zero.
Hence, the flux only passes through the surface are ABCD.
The net flux is ϕnet=ϕABCD=E0a⋅a2
Using Gauss's law, ϵ0qen=E0a3
⇒qen=E0ϵ0a3
=4×104×9×10−12×8×10−6
=288×10−14C
Hence, the value of Q=288.