d
The given figure shows six equal amount of charges, \(q\), at the vertices of a regular hexagon.

Where, Charge, \(q=5\, \mu\, C=5 \times 10^{-6} \,C\)

Side of the hexagon, \(l= AB = BC = CD = DE = EF = FA =10 \,cm\)

Distance of each vertex from centre \(O , d =10\, cm\)

Electric potential at point \(O\)

\(V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{6 \times q}{d}\)

Where,

Where, \(\varepsilon_{0}=\) Permittivity of free space and \(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}\, Nm ^{2} \,C ^{-2}\)

\(\therefore V=\frac{9 \times 10^{9} \times 6 \times 5 \times 10^{-6}}{0.1}=2.7 \times 10^{6}\, V\)

Therefore, the potential at the centre of the hexagon is \(2.7 \times 10^{6} \;V\)