JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Electrostatics

By Gauss's law the enclosed charge is $q={\epsilon }_0\oint \vec{E}\cdot d\vec{A}$, obtained by adding the flux through the six faces of the cuboid, which sits with one corner at the origin and edges $1,2,3$ m along $x,y,z$. For $\vec{E}=2x^2\hat{i}-4y\hat{j}+6\hat{k}$, the faces at $x=0$ and $y=0$ contribute nothing, and the constant $6\hat{k}$ term gives equal and opposite flux on the two $z$-faces. The face at $x=1$ has $E_x=2$ over area $2\times 3$, contributing $+12$, while the face at $y=2$ has $E_y=-8$ over area $1\times 3$, contributing $-24$, so the net flux is $12-24=-12$. Hence $q=-12{\epsilon }_0$, and writing it as $n{\epsilon }_0$ fixes $n=12$.