Answer: 71

Solution:

Substituting, $z = w^{11} = (z^{13})^{11} = z^{143},$ and $z \ne 0,$ so $z^{142} = 1.$ Conversely, any $142$nd root of unity $z$ works with $w = z^{13},$ since then $w^{11} = z^{143} = z.$

Hence $z = \cos\frac{2k\pi}{142} + i\sin\frac{2k\pi}{142}$ for some integer $k,$ and the imaginary part of $z$ is $\sin\frac{k\pi}{71}.$ Since $71$ is prime, for every $k$ with $1 \le k \le 70$ the fraction $\frac{k}{71}$ is already in lowest terms, matching the required form $\sin\left(\frac{m\pi}{n}\right)$ with $m \lt n.$ Thus $n = 71.$