The $n$-gon’s interior angle, the $60^\circ$ angle, and $\angle BA_1A_n$ together fill $360^\circ$ at $A_1$

Setting $\angle BA_1A_n$ equal to the interior angle of a regular $m$-gon gives $\frac{1}{m} = \frac{1}{6} - \frac{1}{n},$ so $n - 6$ must divide $36$

Since $B$ is outside the $n$-gon, the angles at $A_1$ — the interior angle $\angle A_nA_1A_2 = \frac{(n-2)180^\circ}{n},$ the equilateral angle $\angle A_2A_1B = 60^\circ,$ and $\angle BA_1A_n$ — fill a full revolution, so $$\begin{aligned} \angle BA_1A_n &= 300^\circ - \frac{(n-2)180^\circ}{n} \\ &= 120^\circ + \frac{360^\circ}{n}. \end{aligned}$$ Also $A_nA_1 = A_1B,$ since both equal the side of the $n$-gon.

For $A_n,$ $A_1,$ $B$ to be consecutive vertices of a regular $m$-gon, this angle must be the $m$-gon’s interior angle: $120^\circ + \frac{360^\circ}{n} = 180^\circ - \frac{360^\circ}{m},$ which simplifies to $\frac{1}{m} = \frac{1}{6} - \frac{1}{n},$ so $m = \frac{6n}{n - 6} = 6 + \frac{36}{n - 6}.$

Thus $n - 6$ must divide $36,$ and the largest choice is $n - 6 = 36,$ i.e. $n = 42$ (with $m = 7$).