If a point charge is placed at the centre of the cube, then flux linked to the surface shaded in figure.

$\frac{Q}{2ε_0}$

The correct answer is Option (2) → $\frac{Q}{6ε_0}$

$\text{Total flux by Gauss's law: } \Phi_{total} = \frac{Q}{\varepsilon_0}$

Since charge $Q$ is at the center of the cube, the flux is equally shared among the 6 faces.

Flux through one face:

$\;\;\; \Phi_{face} = \frac{Q}{6\varepsilon_0}$

In the figure, 3 faces are shaded. So flux through shaded region:

$\;\;\; \Phi_{shaded} = 3 \times \frac{Q}{6\varepsilon_0} = \frac{Q}{2\varepsilon_0}$