Answer
$ U=(4+\sqrt 2)\frac{KQ^2}{a}$
Work Step by Step
We can find the required potential energy of the system as follows:
$ U=K[\frac{Q^2}{r_{12}}+\frac{Q^2}{r_{13}}+\frac{Q^2}{r_{14}}+\frac{Q^2}{r_{23}}+\frac{Q^2}{r_{24}}+\frac{Q^2}{r_{34}}]$
$ U=KQ^2[\frac{1}{r_{12}}+\frac{1}{r_{13}}+\frac{1}{r_{14}}+\frac{1}{r_{23}}+\frac{1}{r_{24}}+\frac{1}{r_{34}}]$
$ U=KQ^2[\frac{1}{a}+\frac{1}{\sqrt{a^2+a^2}}+\frac{1}{a}+\frac{1}{a}+\frac{1}{\sqrt{a^2+a^2}}+\frac{1}{a}]$
$ U=\frac{KQ^2}{a}[4+\frac{2}{\sqrt2}]$
$ U=(4+\sqrt 2)\frac{KQ^2}{a}$
