Step 1: Recall what work means here.
Bringing a charge $q_0$ from infinity to a point where the potential is $V$ costs work equal to its change in potential energy, $W = q_0 V$. The whole problem therefore reduces to finding the electric potential at the centre of the hexagon.
Step 2: Note the key geometric fact about a regular hexagon.
In a regular hexagon of side $R$, the centre is equidistant from every vertex, and that distance (the circumradius) equals the side itself, $R$. So each of the five fixed charges sits at the same distance $R$ from the centre.
Step 3: Use the superposition of potentials.
Potential is a scalar, so contributions just add algebraically. Each charge contributes $\dfrac{1}{4\pi\varepsilon_0}\dfrac{(\pm q)}{R}$, with the sign of the charge carried along.
Step 4: Add the signed charges first.
There are three $+q$ and two $-q$ charges, so the algebraic sum of charges is \[ 3(+q) + 2(-q) = +q \] Because all distances are equal to $R$, the total potential is \[ V = \frac{1}{4\pi\varepsilon_0}\frac{(+q)}{R} \]
Step 5: Bring in the test charge.
The charge moved from infinity is $q_0 = -q$. Therefore \[ W = q_0 V = (-q)\left(\frac{1}{4\pi\varepsilon_0}\frac{q}{R}\right) \]
Step 6: Simplify to the final form.
\[ W = \frac{1}{4\pi\varepsilon_0}\left(\frac{-q^2}{R}\right) \] The negative sign tells us the configuration is attractive, matching option (A).
\[ \boxed{W = \frac{1}{4\pi\varepsilon_0}\left(\frac{-q^2}{R}\right)} \]