## Task $9 / 82$

The volume and surface area of a sphere are numerically equal. Under this condition, what is the surface area of a cube inscribed in the sphere?
If $r$ is the radius of the sphere, then according to the given condition

$$
\frac{4}{3} \pi r^{3}=4 \pi r^{2} \quad ; \quad r=3 \text{LE}
$$

(where LE denotes the unit of length). For the side length $a$ of the cube inscribed in the sphere with radius $r$, we have

$$
a=\frac{2}{3} \sqrt{3} r=2 \sqrt{3}
$$

(this follows immediately from the fact that the space diagonal of the cube is equal to the diameter of the sphere: $\sqrt{3} a=d=2 r$ ).

Thus, for the surface area $A$ of the cube, we have: $A=6 a^{2}=6 \cdot\left(2 \sqrt{3}\right)^{2}=72 \text{LE}^{2}$.