The figure below represents the three point charges forming
equilateral triangle.

The energy system of three charges,

$U=3\left(\frac{k{q}_1{q}_2}{r}\right)$

$U=3\times \frac{k\left(2\times {10}^{-3}\times 2\times {10}^{-3}\right)}{{\displaystyle \frac{1}{2}}}$

$U=24\mathrm{k}\times {10}^{-6} \mathrm{J}$

When one charge moves to the midpoint of the line joining the
other two charges,

The energy of the system,

$U^{\text{'}}=\frac{k\left(2\times 2\times {10}^{-6}\right)}{(1/4)}\times 2+\frac{k\left(2\times 2\times {10}^{-6}\right)}{(1/2)}$

$U^{\text{'}}=40\mathrm{k}\times {10}^{-6} \mathrm{J}$

The net energy,

$\mathrm{\Delta }U=U^{\text{'}}-U$

$\mathrm{\Delta }U=(40-24)k\times {10}^{-6}$

$\mathrm{\Delta }U=16\times 9\times {10}^3 \mathrm{J}$

The energy supplied to the system at the rate of $2 \mathrm{kW}$,

$2 \mathrm{kW}\times t=16\times 9\times {10}^3$

$2\times {10}^3\times t=16\times 9\times {10}^3$

$t=72 \mathrm{s}$