I would guess it's "...find the least $ m$ such that $ \frac{F_2}{F_1} < m$..." because there is no such $ m$ as it is stated. Areal coordinates: any point on $ w_a$ has coordinates $ \left(\tfrac{x}{x+b+c}, \tfrac{b}{x+b+c}, \tfrac{c}{x+b+c}\right)$, where $ x$ is positive; any point on $ m_b$ has coordinates $ \left(\tfrac{1}{2+y}, \tfrac{y}{2+y}, \tfrac{1}{2+y}\right)$, where $ y$ is positive. Using this we can find that the intersection $ w_a \cap m_b$ has areal coordinates $ \left(\tfrac{c}{b+2c}, \tfrac{b}{b+2c}, \tfrac{c}{b+2c}\right)$. Similarly, $ w_b \cap m_c$ has areal coordinates $ \left(\tfrac{a}{c+2a}, \tfrac{a}{c+2a}, \tfrac{c}{c+2a}\right)$ and $ w_c \cap m_a$ has areal coordinates $ \left(\tfrac{a}{a+2b}, \tfrac{b}{a+2b}, \tfrac{b}{a+2b}\right)$. Hence \[ \frac{F_2}{F_1} = \left|\begin{matrix} \tfrac{c}{b+2c} & \tfrac{b}{b+2c} & \tfrac{c}{b+2c} \\ \tfrac{a}{c+2a} & \tfrac{a}{c+2a} & \tfrac{c}{c+2a} \\ \tfrac{a}{a+2b} & \tfrac{b}{a+2b} & \tfrac{b}{a+2b} \end{matrix}\right| = \frac{ab^2+bc^2+ca^2-3abc}{(a+2b)(b+2c)(c+2a)}\] I would guess that the maximum of the above expression occurs when $ a,b=x, c\to 2x$, which gives $ m = \tfrac{1}{60}$? Otherwise, it is just tedious taking of derivatives.

Actually, $ a,b=x, c\to 0$ gives $m=\frac{1}{6}$, which turns out to be the maximum.