A charge Q is distributed uniformly over a rod of length l . A...

By Gauss's law, the electric flux through the entire surface of the cube is given by $\Phi = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$.

To minimize the flux, the charge enclosed by the cube must be minimized.

The rod has a uniform linear charge density $\lambda = \dfrac{Q}{l}$.

The enclosed charge is $q_{\text{enclosed}} = \lambda l_{\text{enclosed}}$, where $l_{\text{enclosed}}$ is the length of the rod inside the cube.

One end of the rod is fixed at the centre of the cube. To minimize the length of the rod inside the cube, the rod must be oriented along the shortest path from the centre to the surface of the cube.

The shortest distance from the centre of a cube of edge $l$ to its surface is the perpendicular distance to any of its faces, which is $\dfrac{l}{2}$.

Thus, the minimum length of the rod inside the cube is $l_{\text{enclosed}} = \dfrac{l}{2}$.

The minimum enclosed charge is $q_{\text{enclosed}} = \left(\dfrac{Q}{l}\right) \left(\dfrac{l}{2}\right) = \dfrac{Q}{2}$.

Therefore, the minimum possible flux through the cube is \Phi_{\text\min} = \dfrac{Q}{2\varepsilon_0}.