Among the equal volume equidistant surfaces, which one has the smallest surface area?
Let the edges meeting at one vertex be denoted by $x, y, z$. Since the volume $x y z$ is constant, then $x^{2} y^{2} z^{2}=x y \cdot y z \cdot z x$ is also constant. Therefore, half of the surface area: $x y+y z+z x$ is minimal when $x y=y z=z x$, which means $x=y=z$. Thus, the surface area of the cube is the smallest.

Number of solutions: 23.