Answer
$2.1\times 10^{-25}~J~~$ of work is required to bring the $+6e$ charge to the center.
Work Step by Step
We can write a general expression for the change in electric potential energy of the system when we bring in the particle of charge $+6e$:
$\Delta U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{(+6e)(q_i)}{r_i}$
We can find the change in electric potential energy of this system of charged particles:
$\Delta U = \frac{1}{4\pi~\epsilon_0}~[\frac{(+6e)(+5e-2e)}{0.020~m}+\frac{(+6e)(-4e+4e)}{\sqrt{2}~(0.020~m)}]$
$\Delta U = \frac{1}{4\pi~\epsilon_0}~\frac{(+6e)(+3e)}{0.020~m}$
$\Delta U = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{(6)(1.6\times 10^{-19}~C)(3)(1.6\times 10^{-19}~C)}{0.020~m}$
$\Delta U = 2.1\times 10^{-25}~J$
We can find the work that is required to bring the $+6e$ charge to the center:
$W = \Delta U = 2.1\times 10^{-25}~J$
$2.1\times 10^{-25}~J~~$ of work is required to bring the $+6e$ charge to the center.