The surface area of the circumscribed sphere of a cube $K_{1}$ is twice as large as the surface area of the inscribed sphere of a cube $K_{2}$. Let $V_{1}$ denote the volume of the inscribed sphere of cube $K_{1}$, and $V_{2}$ the volume of the circumscribed sphere of cube $K_{2}$. What is the ratio $\frac{V_{1}}{V_{2}}$?
I. solution. We know that the length of the radius of the sphere circumscribed around a cube is equal to half the length of the cube's space diagonal, and the radius of the inscribed sphere is equal to half the length of the cube's edge. Let $a$ be the edge of cube $K_{1}$, the radius of its circumscribed sphere $r_{1}$, its surface area $F_{1}$; the volume of the inscribed sphere $V_{1}$.

$$
r_{1}=\frac{a \sqrt{3}}{2}, \quad F_{1}=4\left(\frac{a \sqrt{3}}{2}\right)^{2} \pi, \quad V_{1}=\frac{4\left(\frac{a}{2}\right)^{3} \pi}{3}
$$

The edge of cube $K_{2}$ is $b$, the radius of its inscribed sphere $r_{2}$, its surface area $F_{2}$, the volume of its circumscribed sphere $V_{2}$.

$$
r_{2}=\frac{b}{2}, \quad F_{2}=4\left(\frac{b}{2}\right)^{2} \pi, \quad V_{2}=\frac{4\left(\frac{b \sqrt{3}}{2}\right)^{3} \pi}{3}
$$

We know that $F_{1}=2 F_{2}$, that is, $4 \pi\left(\frac{a \sqrt{3}}{2}\right)^{2}=2 \cdot 4 \pi\left(\frac{b}{2}\right)^{2}$. From this, after simplification and rearrangement, we get $a=b \sqrt{\frac{2}{3}}$.

The ratio of the sought volumes is:

$$
\frac{V_{1}}{V_{2}}=\frac{\frac{4\left(\frac{a}{2}\right)^{3} \pi}{3}}{\frac{4\left(\frac{b \sqrt{3}}{2}\right)^{3} \pi}{3}}
$$

Perform the simplifications and substitute $a=b \sqrt{\frac{2}{3}}$, we get

$$
\frac{V_{1}}{V_{2}}=\frac{\left(\sqrt{\frac{2}{3}}\right)^{3}}{(\sqrt{3})^{3}}=\frac{2 \sqrt{2}}{27} \approx 0.1048
$$

II. solution. Let $f\left(K_{i}\right)$ and $K_{i}(f)$ denote the surface area of the circumscribed and inscribed spheres of the $i$-th cube $(i=1,2)$, and $v\left(K_{i}\right)$ and $K_{i}(v)$ the volumes of these spheres.

The circumscribed and inscribed spheres of a cube are similar, the ratio of similarity being the ratio of the diameters, or the ratio of the cube's edge and space diagonal, $1: \sqrt{3}$. The ratio of the surface areas, and the volumes, of similar bodies is the square, or the cube, of the ratio of similarity; thus $\frac{f\left(K_{i}\right)}{K_{i}(f)}=(\sqrt{3})^{2}=3$ and $\frac{v\left(K_{i}\right)}{K_{i}(v)}=(\sqrt{3})^{3}=3 \sqrt{3}$.

According to the condition, $\frac{f\left(K_{1}\right)}{K_{2}(f)}=2$. From this,

$$
\frac{f\left(K_{1}\right)}{f\left(K_{2}\right)}=\frac{f\left(K_{1}\right)}{K_{2}(f)} \cdot \frac{K_{2}(f)}{f\left(K_{2}\right)}=2 \cdot\left(\frac{1}{\sqrt{3}}\right)^{2}=\frac{2}{3}
$$

is the ratio of the surface areas of the circumscribed spheres of $K_{1}$ and $K_{2}$. This is also the ratio of the surface areas of the cubes, the square of the ratio of similarity of the cubes. The cube of this ratio is also the ratio of the volumes of the two cubes - and thus the ratio of the volumes of the two inscribed, and circumscribed, spheres.

The ratio in question in the problem is therefore

$$
\frac{V_{1}}{V_{2}}=\frac{K_{1}(v)}{v\left(K_{2}\right)}=\frac{K_{1}(v)}{K_{2}(v)} \cdot \frac{K_{2}(v)}{v\left(K_{2}\right)}=\left(\sqrt{\frac{2}{3}}\right)^{3} \cdot \frac{1}{3 \sqrt{3}}=\frac{2 \sqrt{2}}{27}
$$