d
(d) Length of the diagonal of a cube having each side b is \(\sqrt 3 \,b.\) So distance of centre of cube from each vertex is \(\frac{{\sqrt 3 \,b}}{2}.\)
Hence potential energy of the given system of charge is
\(U = 8 \times \left\{ {\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{( - q)\,(q)}}{{\sqrt 3 \,b/2}}} \right\} = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}\)