= Solution
For a regular hexagon of side $a$, the perpendicular centre-to-side distance is $d=\sqrt3a/2$. Each side subtends endpoint <angles> with $\cos\beta=1/2$ and $\cos\alpha=-1/2$. Its centre field is consequently $\mu_0I/(4\pi d)$.
All six contributions have the same normal direction, so
$$
\boxed{B_{\rm centre}=6\frac{\mu_0I}{4\pi d}=\frac{\sqrt3\,\mu_0I}{\pi a}.}
$$
The direction follows the current's <right-hand rule>. Equal side lengths alone would not fix an arbitrary distorted hexagon's field; the usual regular-coil geometry is used here.
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