Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is (assume x >>a)

Two opposite charges situated at opposite vertices forms a dipole and point lies on bisector line of dipole for one dipole.

$\overrightarrow{\mathrm{E}}=\frac{-\mathrm{K}\overrightarrow{\mathrm{p}}}{{\mathrm{x}}^3}$

Direction of field due to these three dipoles is in a horizontal plane at distance x from plane of hexagon and parallel to it, mutually having angle 60° with each other.

${\mathrm{E}}_0=\mathrm{E}+\mathrm{E}=2\mathrm{E}=2\frac{\mathrm{Q}\left(2\mathrm{a}\right)}{4{\mathrm{\pi \epsilon }}_0\cdot {\mathrm{x}}^3}=\frac{\mathrm{Qa}}{{\mathrm{\pi \epsilon }}_0{\mathrm{x}}^3}$