1. Flux through face due to charge at the centre of the cube

A charge $q$ is placed at the centre $O$ of a cube.

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

Since the charge is at the centre, by symmetry this flux is equally distributed through the 6 faces.

So flux through face $ABCD$ due to the central charge is $\Phi_1=\frac{1}{6}\cdot \frac{q}{\varepsilon_0}=\frac{q}{6\varepsilon_0}.$

2. Flux through face $ABCD$ due to the second charge

Another same charge $q$ is placed at a distance $L$ from $O$.

Here the question does not specify the direction of this second charge relative to the cube. Therefore, its contribution to flux through face $ABCD$ cannot be uniquely determined from the given information unless some special geometry is intended.

Also, the options given are dimensionally inconsistent: $\frac{q}{\pi\varepsilon_0 L}$ has dimensions of electric field $\times$ area mismatch, whereas electric flux should have units of $\frac{q}{\varepsilon_0}.$ So options A, C, D are not dimensionally valid expressions for flux.

3. Check the given options

• A: $\dfrac{q}{4\pi\varepsilon_0 L}$ → dimensionally incorrect for flux.
• B: zero
• C: $\dfrac{q}{2\pi\varepsilon_0 L}$ → dimensionally incorrect.
• D: $\dfrac{q}{3\pi\varepsilon_0 L}$ → dimensionally incorrect.

Since the central charge alone gives nonzero flux $\frac{q}{6\varepsilon_0},$ flux through $ABCD$ cannot be zero.

Hence option B is physically not correct either.

4. Conclusion

The physically correct flux through face $ABCD$ due to the charge at the centre is $\boxed{\frac{q}{6\varepsilon_0}}$ if only the central charge contributes symmetrically.

Because the position of the second charge is not fully specified, the problem as written is ambiguous. Therefore the stored answer $\text{B}$ cannot be accepted as correct.