1. Use Gauss's law

The electric flux through a closed surface is

$\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

So we only need the net charge enclosed by the cube.

The cube is bounded by

$x=\pm \frac a2,\quad y=\pm \frac a2,\quad z=\pm \frac a2.$

Thus, any charge lying inside this region contributes fully; if only part of a continuous distribution lies inside, only that part contributes.

2. Charge on the disk

• Disk radius: $\dfrac a4$
• Total charge on disk: $6\,\text{C}$
• Centre of disk: $\left(-\dfrac a2,0,0\right)$
• Disk lies in the $xy$-plane.

The plane $x=-\dfrac a2$ is a face of the cube, and the disk is centered exactly on this face. Since the disk has radius $a/4$, all its points satisfy

$x=-\frac a2, \qquad -\frac a4 \le y \le \frac a4,$

so the entire disk lies on the boundary surface of the cube.

For flux by Gauss's law, a charge distributed symmetrically on the boundary contributes half its charge as enclosed charge.

Hence enclosed charge from disk:

$Q_{\text{disk,encl}} = \frac{6}{2} = 3\,\text{C}.$

3. Charge on the rod

• Rod length: $a$
• Total charge: $8\,\text{C}$
• Rod extends along x-axis from

$x=\frac a4 \text{ to } x=\frac{5a}{4}.$

The cube includes x-values from $-a/2$ to $a/2$.

So only the part of the rod from

$x=\frac a4 \text{ to } x=\frac a2$

lies inside the cube.

This part has length

$\frac a2 - \frac a4 = \frac a4.$

Since the full rod length is $a$, the fraction inside is

$\frac{a/4}{a} = \frac14.$

Therefore enclosed rod charge is

$Q_{\text{rod,encl}} = 8 \times \frac14 = 2\,\text{C}.$

4. Point charge $-7\,\text{C}$

This charge is at

$\left(\frac a4,-\frac a4,0\right).$

Check whether it lies inside the cube:

$-\frac a2 \le \frac a4 \le \frac a2, \qquad -\frac a2 \le -\frac a4 \le \frac a2, \qquad -\frac a2 \le 0 \le \frac a2.$

So it is inside the cube.

Hence contribution:

$Q_1=-7\,\text{C}.$

5. Point charge $3\,\text{C}$

This charge is at

$\left(-\frac{3a}{4},\frac{3a}{4},0\right).$

But

$-\frac{3a}{4} < -\frac a2, \qquad \frac{3a}{4} > \frac a2,$

so this point lies outside the cube.

Hence contribution:

$Q_2=0.$

6. Net enclosed charge

Now add all enclosed contributions:

$Q_{\text{enclosed}} = 3 + 2 - 7 = -2\,\text{C}.$

7. Electric flux through the cube

By Gauss's law,

$\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} = \frac{-2}{\varepsilon_0}.$

So the required flux is

$\boxed{\frac{-2\text{ C}}{\varepsilon_0}}.$

This matches Option A.

8. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

So they agree.