Answer
The amount of work required to bring the charge $Q$ to this position is zero.
Work Step by Step
We can write a general expression for the change in electric potential energy of the system when we bring in the particle of charge $Q$:
$\Delta U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{Q~q_i}{r_i}$
We can find the change in electric potential energy of this system of charged particles:
$\Delta U = \frac{1}{4\pi~\epsilon_0}~(\frac{Q~q_1}{2.00~d}+\frac{Q~q_2}{d})$
$\Delta U = \frac{1}{4\pi~\epsilon_0}~(\frac{Q~q_1}{2.00~d}-\frac{Q~q_1}{2d})$
$\Delta U = \frac{1}{4\pi~\epsilon_0}~(0)$
$\Delta U = 0$
Since the required work is equal to $\Delta U$, the amount of work required to bring the charge $Q$ to this position is zero.
