\({V_O} = 4\,\left( {\frac{Q}{{4\pi {\varepsilon _0}(a/\sqrt 2 )}}} \right)\)
Work done in shifting \((-Q)\) charge from centre to infinity \(W = - \,Q({V_\infty } - {V_O}) = Q{V_0}\)\( = \frac{{4\sqrt 2 \,{Q^2}}}{{4\pi {\varepsilon _0}a}}\)\( = \frac{{\sqrt 2 {Q^2}}}{{\pi {\varepsilon _0}a}}\)