As shown in figure, a cuboid lies in a region with electric field E...

Given:

\overset{\to}{E} = 2{x}^{2}\overset{^}{i} - 4y\overset{^}{j} + 6\overset{^}{k}

For top and bottom faces, as value of $x = 0 \text{ and } y = 0$, electric field becomes constant and therefore net flux through these two faces will be zero(${ϕ}_{z} = 0$).

Now, ${E}_{x} = 2{x}^{2}$, ${E}_{y} = - 4y$ and ${E}_{z} = 6$.

Change in net flux through face at x=1 and at x=0 parallel to y-z plane can written as, ${ϕ}_{x} = \left(2 \times {1}^{2}\right) \times \left(2 \times 3\right) - 0 = 12$.

Similarly, net flux through face at y=2 and y=0 parallel to x-z plane can be written as, ${ϕ}_{y} = - \left(4 \times 2\right) \times \left(1 \times 3\right) - 0 = - 24$

${ϕ}_{net} = 12 - 24 = - 12$

Using Gauss's law,

${ϕ}_{net} = \dfrac{q}{{ε}_{0}}$

$⇒ \left|q\right| = 12{ε}_{0}$