Given:
E=2x2i^−4yj^+6k^

For top and bottom faces, as value of x=0&y=0, electric field becomes constant and therefore net flux through these two faces will be zero(ϕz=0).
Now, Ex=2x2, Ey=−4y and Ez=6.
Change in net flux through face at x=1 and at x=0 parallel to y-z plane can written as, ϕx=(2×12)×(2×3)−0=12.
Similarly, net flux through face at y=2 and y=0 parallel to x-z plane can be written as, ϕy=−(4×2)×(1×3)−0=−24
ϕnet=12−24=−12
Using Gauss's law,
ϕnet=ϵ0q
⇒∣q∣=12ϵ0
