Answer
$W=-24J$
Work Step by Step
We can find the initial potential energy of the charges as:
$U_i=\frac{Kq_1q_2}{b}+\frac{Kq_1q_3}{a}+\frac{Kq_2q_3}{a}$
We plug in the known values to obtain:
$U_i=\frac{9\times 10^9\times 10\times 10^{-6}\times(-20\times 10^{-6})}{0.06}+\frac{9\times 10^9\times 10\times 10^{-6}\times(30\times 10^{-6})}{0.10}+\frac{9\times 10^9\times (-20\times 10^{-6})\times(30\times 10^{-6})}{0.10}=-57J$
Now the final potential energy is
$U_f=\frac{Kq_1q_2}{a}+\frac{Kq_1q_3}{a}+\frac{Kq_2q_3}{b}$
We plug in the known values to obtain:
$U_f=\frac{9\times 10^9\times 10\times 10^{-6}\times(-20\times 10^{-6})}{0.10}+\frac{9\times 10^9\times 10\times 10^{-6}\times(30\times 10^{-6})}{0.10}+\frac{9\times 10^9\times (-20\times 10^{-6})\times(30\times 10^{-6})}{0.06}=-81J$
Now work done $W$ can be determined as:
$W=U_f-U_i=-81-(-57)=-24J$