This problem can be solved in several different ways. Here, we present a solution utilizing the Menelaus Theorem. Other solutions are contained in the book %%HREF%%Geometry Techniques%%Home/35-books/94-book-geometry-techniques%%. Join $BE$ and let its intersection point with $ABC$ be $P$.

Apply Menelaus' theorem on $\triangle{CPE}$ with respect to line $BMN$: \begin{equation} \frac{CM}{MP}\cdot\frac{PB}{BE}\cdot\frac{EN}{NC}=-1 \end{equation} We note that: - $\frac{CM}{MP}=\dfrac{1-r}{r-\frac{1}{2}}$ - $\frac{PB}{BE}=-\dfrac{1}{4}$ (because $BP=AB\sin{30^\circ}$) - $\frac{BN}{BC}=\dfrac{1-r}{r}$ Setting $(i), (ii), (iii)$ to the previous equation leads to: $$\frac{1-r}{r-\frac{1}{2}}\cdot(-\frac{1}{4})\cdot\frac{1-r}{r}=-1 \implies r=\boxed{\frac{\sqrt{3}}{3}}$$