Answer
The same amount 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.
Note that the amount of work required depends on the initial position and the final position, but it does not depend on the specific path taken between the two positions.
Therefore, the same amount of work is required.
