Polya noted the solution for Problem 8 (Axes of a Cube) in How to Solve it:
... there are three kinds of axes:
(1) 4 axes, each through two opposite vertices ...
(2) 6 axes, each through the mid-points of two opposite edges ...
(3) 3 axes, each through the center of two opposite faces ...
... The desired average is
$$\frac{4\sqrt{3}+6\sqrt{2}+3}{13}=1.416$$
(This problem may be useful in preparing the reader for the study of crystallography. For the reader sufficiently advanced in the integral calculus it may be observed that the average computed is a fairly good approximation to the “average width” of the cube, which is, in fact, 3/2=1.5)
I guessed two means of "average width" here.
Solution 1
If the "average width" means the diameter that is equivalent to a sphere whith same volume, it should be calculated as: $$\frac{4\pi r^3}{3}=a^3$$ so we get: $$2r=\sqrt[3]{\frac{6}{\pi}}a\approx1.24a$$
Solution 2
As I discussed with Empy2, if it means the integral over all directions, the solution is more complicated.
Let's use spherical coordinates
$$\begin{cases}x=r\cos\phi\sin\theta\\y=r\sin\phi\sin\theta\\z=r\cos\theta\end{cases}$$
Then the average radius can be defined as:
$$r_{Avg}=\frac{1}{4\pi}\int_\Omega rd\Omega=\frac{1}{4\pi}\int_\phi\int_\theta r\sin\theta d\theta d\phi$$
Here let's choose a cube with edge length 2, centered at the origin, then calculate the average radius (half of width) of one of the symmetric 48 parts of the cube:
Then this part of the cube is a tetrahedron with 4 planes:
- OAB: y = 0, or φ = 0
- OAC: x = y, or φ = π / 4
- OBC: x = z, or cosφ sinθ = cosθ
- ABC: z = 1, or r cosθ = 1
So we have:
$$\begin{cases}\phi_{OAB}=0\\\phi_{OAC}=\pi/4\\\theta_{OA}=0\\\theta_{OBC}=\text{arccot}\cos\phi\\r_{ABC}=1/\cos\theta\end{cases}$$
Because of the symmetry, we can just integrate the average radius of the tetrahedron:
$$r_{Avg}=\frac{48}{4\pi}\int_{\phi_{OAB}}^{\phi_{OAC}}\int_{\theta_{OA}}^{\theta_{OBC}}r_{ABC}\sin\theta d\theta d\phi=\frac{12}\pi\int_0^{\pi/4}\int_0^{\text{arccot}\cos\phi}\frac{\sin\theta}{\cos\theta}d\theta d\phi$$
Let's calculate the integral:
$$\int_0^{\text{arccot}\cos\phi}\frac{\sin\theta}{\cos\theta}d\theta=-\ln|\cos\theta|_0^{\text{arccot}\cos\phi}=\ln\sqrt{\cos^2\phi+1}-\ln\cos\phi$$
Although the antiderivative of this function doesn't look analytic, we can do numeric integral:
$$r_{Avg}=\frac{12}\pi\int_0^{\pi/4}(\ln\sqrt{\cos^2\phi+1}-\ln\cos\phi)d\phi\approx1.2214$$
which is close to the solution 1 (volume definition) but far different from Polya's note.
The width is the distance between parallel planes that are as close together as possible without intersecting the inside of the cube. . They will normally go through opposite corners of the cube.
You have used the line through the centre of the cube, that end where the cube ends. The planes perpendicular to the line through its endpoints will usually intersect the cube, and the planes that define the width will be further apart.