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 $2.7 \times 10^{6} \;V$