11. The faces of a hexahedron and the faces of a regular octahedron are all equilateral triangles with side length $a$. The ratio of the radii of the inscribed spheres of these two polyhedra is a reduced fraction $\frac{m}{n}$. Then, the product $m \cdot n$ is $\qquad$.
For a hexahedron, $B_{1} F_{1}$ is the height of a regular tetrahedron, $B_{1} F_{1}=\frac{\sqrt{6}}{3} a$, $O_{1} H_{1}=\frac{1}{3} B_{1} F_{1}=\frac{\sqrt{6}}{9} a ;$
For a regular octahedron, $O_{2} P_{2}=\frac{\sqrt{2}}{2} a$, $O_{2} H_{2} \cdot P_{2} E_{2}=O_{2} E_{2} \cdot O_{2} P_{2}$
$\Rightarrow O_{2} H_{2}=\frac{O_{2} E_{2} \cdot O_{2} P_{2}}{P_{2} E_{2}}=\frac{\sqrt{6}}{6} a$.
Thus $\frac{m}{n}=\frac{2}{3} \Rightarrow m \cdot n=6$.