Place the center at the origin; by symmetry the smaller hexagon is regular and concentric, so it suffices to find the distance from the center to one of its vertices

One vertex of the small hexagon is the intersection of lines $AH$ and $FG;$ with circumradius $1$ it is $\left(\frac{5}{7}, \frac{\sqrt{3}}{7}\right)$

Center the hexagon at the origin with circumradius $1:$ $A = (1, 0),$ $B = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $C = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ and $F = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).$ Then $H = \left(0, \frac{\sqrt{3}}{2}\right)$ and $G = \left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).$ Rotation by $60^\circ$ permutes the six segments, so the smaller hexagon is regular and concentric, and the area ratio is the square of the ratio of distances from the center to a vertex.

One vertex is the intersection of $AH$ and $FG.$ Line $AH$ is $x + \frac{2}{\sqrt{3}}\,y = 1,$ and line $FG$ is $y = 3\sqrt{3}\,x - 2\sqrt{3}.$ Substituting gives $x + 6x - 4 = 1,$ so $x = \frac{5}{7}$ and $y = \frac{\sqrt{3}}{7}.$

That vertex has squared distance $\frac{25}{49} + \frac{3}{49} = \frac{4}{7}$ from the center, while $A$ is at distance $1.$ The ratio of areas is $\frac{4}{7},$ and $m + n = 4 + 7 = 11.$