1. Interpret the geometry

The cuboid has dimensions $2L \times 2L \times L.$

So its faces are:

• two faces of area $4L^2$,
• four faces of area $2L^2$.

A charge $q$ is placed at the center of the face $S$ whose area is $4L^2$. We need the electric flux through the opposite face (the parallel face of the same area $4L^2$).

2. Use symmetry by extending the cuboid

Since the charge is at the center of one face, imagine attaching an identical cuboid to the given one on the other side of that face. Then the given face becomes an internal plane, and the charge lies at the center of the resulting larger cuboid.

The new larger cuboid has dimensions $2L \times 2L \times 2L,$ which is actually a cube of side $2L$.

The charge $q$ is now at the center of the cube.

3. Total flux through the cube

By Gauss's law, total electric flux through the closed cube is $\Phi_{\text{total}} = \frac{q}{\varepsilon_0}.$

Since the charge is at the center of the cube, by symmetry this flux is equally distributed among the 6 faces.

Hence flux through each face of the cube is $\Phi_{\text{each face}} = \frac{1}{6}\cdot \frac{q}{\varepsilon_0} = \frac{q}{6\varepsilon_0}.$

4. Relate this to the required face

The opposite face of the original cuboid is exactly one face of this larger cube. Therefore, the flux through the opposite surface is $\boxed{\frac{q}{6\varepsilon_0}}.$

5. Check options

• A: $\dfrac{q}{2\varepsilon_0}$ ❌
• B: $\dfrac{q}{3\varepsilon_0}$ ❌
• C: $\dfrac{q}{12\varepsilon_0}$ ❌
• D: $\dfrac{q}{6\varepsilon_0}$ ✅

So the correct option is D.