Answer
$2.5~J~~$ of work is required.
Work Step by Step
In part (a), we found that $V_A = 6.0\times 10^4~V$
In part (b), we found that $V_B = -7.8\times 10^5~V$
We can find the work required to move the charge $q_3$ from B to A:
$Work = \Delta U$
$Work = U_A-U_B$
$Work = q_3~V_A-q_3~V_B$
$Work = q_3~(V_A-V_B)$
$Work = (3.0\times 10^{-6}~C)~[6.0\times 10^4~V-(-7.8\times 10^5~V)]$
$Work = (3.0\times 10^{-6}~C)~(8.4\times 10^5~V)$
$Work = 2.5~J$
$2.5~J~~$ of work is required.
