There are eight corners of a cube and in each corner there is a charge of (–q). At the centre of the corner there is a charge of (+q). Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.

Diagonal of the cube = $\sqrt {{b^2} + {b^2} + {b^2}} = \sqrt 3 b$

$\therefore$ $d = \sqrt 3 b/2$

Now, electric potential energy of the charge (+q) due to a charge (–q) at one corner

U $= {{{q_1}{q_2}} \over {4\pi {\varepsilon _0}r}} = {{\left( { + q} \right) \times \left( { - q} \right)} \over {4\pi {\varepsilon _0}\left( {\sqrt 3 b/2} \right)}} = - {{{q^2}} \over {2\pi {\varepsilon _0}\left( {\sqrt 3 b} \right)}}$

$\therefore$ Total electric potential energy due to all the eight identical charges
= $8U = - {{8{q^2}} \over {2\pi {\varepsilon _0}\sqrt 3 b}} = {{ - 4{q^2}} \over {\sqrt 3 \pi {\varepsilon _0}b}}$