Answer
The work done by our force on the particle of charge $+q$ is $~~0$
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 D:
$U_D = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{4d}+\frac{Q^2}{3d})$
Note that $U_A = U_D$
There is no change in the electric potential energy of the three-particle system.
We can find the work done by our force on the particle of charge $+q$:
$Work = \Delta U = 0$
The work done by our force on the particle of charge $+q$ is $~~0$
