Explanation:
The situation is shown in figure.

In figure, \(AC = BD = \sqrt {{a^2} + {a^2}} = a\sqrt 2 \)
and \(OA = OB = OC = OD = \frac{{a\sqrt 2 }}{2} = \frac{a}{{\sqrt 2 }}\)
The potential at the centre \(O\) due to given charge configuration is
\({V_O} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{q}{{OA}} + \frac{q}{{OB}} + \frac{q}{{OC}} + \frac{q}{{OD}}} \right]\)
\[ = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\begin{array}{*{20}{l}}
{\frac{q}{{(a/\sqrt 2 )}} + \frac{q}{{(a/\sqrt 2 )}} + \frac{q}{{(a/\sqrt 2 )}}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{q}{{(a/\sqrt 2 )}}}
\end{array}} \right]\]
\( = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{4q}}{{(a/\sqrt 2 )}}} \right] = \frac{{\sqrt 2 q}}{{\pi {\varepsilon _0}a}}\)
Work done in carrying a charge \(-q\) from centre \(O\) to infinity is
\({W_{0\infty }} = - q\left( {{V_\infty } - {V_0}} \right)\)
\(\,\,\,\,\,\,\,\,\,\, = - q\left( {0 - \frac{{\sqrt 2 q}}{{\pi {\varepsilon _0}a}}} \right)\quad \left( {\because {V_\infty } = 0} \right)\)
\(\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 {q^2}}}{{\pi {\varepsilon _0}a}}\)