Solution:
First note that the long diagonal of the cube has length $2\sqrt{3}$, so the radius of the sphere is $\sqrt{3}$. Let $x$ be the side length of the smaller cube. Then the distance from the center of the sphere to the far face of the smaller cube is $1 + x$, while the distance from the center of the far face to a vertex lying on the sphere is $\frac{x\sqrt{2}}{2}$. Therefore, the square of the radius is
$$3 = (1 + x)^2 + \frac{x^2}{2},$$
or
$$3x^2 + 4x - 4 = (3x - 2)(x + 2) = 0,$$
so $x = \frac{2}{3}$.