Answer
The magnitude of the work done by our force is equal in all three cases.
Work Step by Step
We can write a general expression for the electric potential energy of a system of charged particles:
$U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i~q_j}{r_{ij}}$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point A:
$U_A = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{4d}+\frac{Q^2}{3d})$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point B:
$U_B = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{2d}+\frac{Q^2}{3d})$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point C:
$U_C = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{2d}+\frac{Q^2}{3d})$
We can find an expression for the electric potential energy of the system when the $+q$ charge is at point D:
$U_D = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{4d}+\frac{Q^2}{3d})$
We can find an expression for the work done by our force if the particle is moved from A to B:
$Work = \Delta U = U_B-U_A = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{2d}-\frac{q~Q}{4d}) = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{4d})$
We can find an expression for the work done by our force if the particle is moved from A to C:
$Work = \Delta U = U_C-U_A = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{2d}-\frac{q~Q}{4d}) = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{4d})$
We can find an expression for the work done by our force if the particle is moved from B to D:
$Work = \Delta U = U_D-U_B = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{4d}-\frac{q~Q}{2d}) = -\frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{4d})$
The magnitude of the work done by our force is equal in all three cases.
