3. 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 $\qquad$.
6
3.【Analysis and Solution】Solution 1 Let the radius of the inscribed sphere of this hexahedron be $r_{1}$. On one hand, the volume of this hexahedron can be calculated by multiplying the radius of the inscribed sphere by the surface area and then dividing by $\frac{1}{3}$. On the other hand, it can also be calculated as the volume of two regular tetrahedrons with edge length $a$. Therefore, we have
$$
\frac{1}{3} \times 6 \times \frac{\sqrt{3}}{4} a^{2} \times r_{1}=2 \times \frac{1}{3} \times \frac{\sqrt{3}}{4} a^{2} \cdot \sqrt{a^{2}-\left(\frac{\sqrt{3} a}{3}\right)^{2}} .
$$

Let the radius of the inscribed sphere of the regular octahedron be $r_{2}$. On one hand, the volume of this regular octahedron can be calculated by multiplying the radius of the inscribed sphere by the surface area and then dividing by $\frac{1}{3}$. On the other hand, it can also be calculated as the volume of two regular square pyramids with edge length $a$. Therefore, we have
84
$$
\frac{1}{3} \times 8 \times \frac{\sqrt{3}}{4} a^{2} \cdot r_{2}=2 \times \frac{1}{3} \times a^{2} \times \sqrt{a^{2}-\left(\frac{\sqrt{2} a}{2}\right)^{2}} .
$$

From equations (1) and (2), we get $\frac{r_{1}}{r_{2}}=\frac{2}{3}$.
Therefore, $\frac{m}{n}=\frac{2}{3}, m \cdot n=6$.
Solution 2 As shown in Figure $\mathrm{a}$, the radius of the inscribed sphere of the hexahedron is the perpendicular distance $O R$ from the center $O$ of the equilateral $\triangle B C D$ to one of the faces $A B D$, and $O R$ is also the altitude from the hypotenuse $A M$ of the right $\triangle A O M$.
$$
\therefore r_{1}=O R=\frac{A O \cdot O M}{A M} \text {. }
$$

But $A M=\frac{\sqrt{3}}{2} a, O M=\frac{\sqrt{3}}{6} a$,
$$
\therefore A O=\frac{\sqrt{6}}{3} a \text {. }
$$

Substituting into (3) gives $r_{1}=\frac{\sqrt{6} a}{9}$.
In Figure $\mathrm{b}$, the center of the inscribed sphere of the regular octahedron is the center $O$ of the square $B C D E$, and the perpendicular distance $O R$ from $O$ to one of the faces $A B C$ is the radius $r_{2}$ of the inscribed sphere of the regular octahedron, and $O R$ is also the altitude from the hypotenuse of the right $\triangle A O M$.
$$
\therefore r_{2}=O R=\frac{A O \times O M}{A M} \text {. }
$$

But $A M=\frac{\sqrt{3}}{2} a, O M=\frac{a}{2}$,
$$
\therefore A O=\frac{\sqrt{2}}{2} a \text {. }
$$

Substituting into (4) gives $r_{2}=\frac{\sqrt{6}}{6} a$.
$$
\begin{array}{l}
\therefore \frac{r_{1}}{r_{2}}=\frac{\frac{\sqrt{6} a}{9}}{\frac{\sqrt{6}}{6} a}=\frac{2}{3} \cdot 2,3 \text { are coprime, } \\
\therefore m=2, n=3, m n=2 \times 3=6 .
\end{array}
$$