12: Use 6 equal-length thin iron rods to weld into a regular tetrahedral frame, ignoring the thickness of the iron rods and welding errors. Let the radius of the largest sphere that this frame can contain be $R_{1}$, and the radius of the smallest sphere that can enclose this frame be $R_{2}$. Then $\frac{R_{1}}{R_{2}}$ equals $\qquad$
12. $\frac{\sqrt{3}}{3}$.

According to the problem, $R_{1}$ is the radius of the edge-tangent sphere of this regular tetrahedron frame, and $R_{2}$ is the radius of the circumscribed sphere. Suppose the edge length of the regular tetrahedron is 1, it is easy to know that the diameter of the edge-tangent sphere is the distance between opposite edges of the regular tetrahedron, which is $\frac{\sqrt{2}}{2}$, so, $R_{1}=\frac{\sqrt{2}}{4}$. Also, the radius of the circumscribed sphere is $R_{2}=\frac{\sqrt{6}}{4}$, so, $\frac{R_{1}}{R_{2}}=\frac{\sqrt{3}}{3}$.