$ \therefore \displaystyle\frac{n}{\sin\displaystyle\frac{\pi}{6}} = \displaystyle\frac{4n\sin C}{\sin B}$

$ \therefore\displaystyle\frac{1}{\displaystyle\frac{1}{2}} =\displaystyle \frac{4\sin C}{\sin( \displaystyle\frac{2\pi}{3}-C)}$

$\therefore2 = \displaystyle\frac{4\sin C}{\displaystyle\frac{\sqrt{3}}{2}\cos C -\displaystyle \frac{1}{2}\sin C}$

$\therefore\begin{cases} 5\sin C = \sqrt{3}\cos C \\ \sin^2 C + \cos^2 C = 1 \end{cases}$

$\therefore\sin C =\displaystyle \frac{\sqrt{21}}{14}$