A disk of radius $\frac{\mathrm{a}}{4}$having a uniformly distributed charge 6 C is placed in the x-y plane with its centre at $\left(-\frac{\mathrm{a}}{2}, 0, 0\right)$. A rod of length ‘a’ carrying a uniformly distributed charge 8C is placed on the x-axis from $\mathrm{x}=\frac{\mathrm{a}}{4}$ to $\mathrm{x}=\frac{5\mathrm{a}}{4}$. Two point charges -7C and 3C are placed at $\left(\frac{\mathrm{a}}{4}, -\frac{\mathrm{a}}{4}, 0\right) \mathrm{and} \left(-\frac{3\mathrm{a}}{4}, \frac{3\mathrm{a}}{4}, 0\right)$ respectively.
Consider a cubical surface formed by six surfaces $\mathrm{x}=\pm \frac{\mathrm{a}}{2}, \mathrm{y}=\pm \frac{\mathrm{a}}{2}, \mathrm{z}=\pm \frac{\mathrm{a}}{2}$. The electric flux through this cubical surface is

From Gauss' Law, we have flux equal to
$\varphi =\frac{{\mathrm{\Sigma Q}}_{\text{enc }}}{{\mathrm{\epsilon }}_0}=\frac{\left(\frac{8\mathrm{C}}{4}\right)-7\mathrm{C}+\left(\frac{6\mathrm{C}}{2}\right)}{{\mathrm{\epsilon }}_0}=-\frac{2\mathrm{C}}{{\mathrm{\epsilon }}_0}$

OR

The disc is placed with its center at (-a/2, 0, 0). Hence we can say that half of the discs are enclosed within the cube. Therefore, the charge on the disc enclosed within the cube is 6C/2= 3C.
The length of the rod is ’a’ with total charge 8C. The surface of the cube lies at a distance of a/2. Therefore the section of the rod enclosed within the cube is a/2-a/4=a/4.Therefore the total charge of the rod enclosed within the cubical surface is 2C. It is to be noted that the charge 3C lies outside the cubical surface. The charge -7C lies within the cubical surface i.e. at (a/4, -a/4, 0).
From the above information, the total charge (Q)enclosed within the surface is,
Q=3C+2C+−7CQ=−2C

$\varphi =-\frac{2\mathrm{C}}{{\mathrm{\epsilon }}_0}$