Answer
The work done by our force is positive.
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:
$U_B = \frac{1}{4\pi~\epsilon_0}~(\frac{q~Q}{d}+\frac{q~Q}{2d}+\frac{Q^2}{3d})$
Note that $U_B \gt U_A$ because $\frac{q~Q}{2d}\gt \frac{q~Q}{4d}$
We can find an expression for the work done by our force:
$Work = \Delta U = U_B-U_A \gt 0$
The work done by our force is positive.
