A positive charge q is situated at a distance d from the centre of...

The net electric field inside a solid conducting cube under electrostatic conditions is zero.

The total electric field at the centre of the cube is the vector sum of the electric field due to the external point charge $q$ and the electric field due to the induced charges on the surface of the cube.

$\vec{E}_{\text{net}} = \vec{E}_{q} + \vec{E}_{\text{induced}} = 0$

$\Rightarrow \vec{E}_{\text{induced}} = -\vec{E}_{q}$

The electric field at the centre of the cube due to the positive point charge $q$ at a distance $d$ is given by:

$|\vec{E}_{q}| = \dfrac{q}{4\pi\varepsilon_0 d^2}$

Since $q$ is a positive charge, $\vec{E}_{q}$ is directed away from the charge $q$.

Therefore, the electric field due to the induced charges on the surface of the cube must have the same magnitude but opposite direction to cancel $\vec{E}_{q}$.

$|\vec{E}_{\text{induced}}| = \dfrac{q}{4\pi\varepsilon_0 d^2}$

The direction of $\vec{E}_{\text{induced}}$ is towards the charge $q$.