As shown in the figure, $O$ is the centre of hexagon $A B C D E F$ of each side $8 \mathrm{~cm}$. As it is a regular hexagon $O A B, O B C$, etc are equilateral triangles.
$$
\begin{aligned}
\therefore \quad O A=O B=O C= & O D=O E=O F=8 \mathrm{~cm} \\
& =8 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
The potential at $O$ is
$$
\begin{aligned}
V & =6 \times \frac{q}{4 \pi \varepsilon_0 r} \\
& =\frac{6 \times 9 \times 10^9 \times 4 \times 10^{-6}}{8 \times 10^{-2}}=2.7 \times 10^6 \mathrm{~V}
\end{aligned}
$$