For the answer to the question above, I'll show the computation below.
V(cube) = s^3
SA(cube with no top) = 5s^2
s^3 = 5s^2
s = 5
V = 125 sq ft = 216,000 sq in
V(cannonball) = (4/3)πr^3
SA(cannonball) = 4πr^3
(4/3)πr^3 = 4πr^2
r = 3
V = 36π sq in
Packing of spheres in a cube is yielding up to 74% with remainder as space between the spheres. It can be achieved by placing the 2nd row of spheres in the crevices between adjoining first row spheres.
216,000 / 36π = 6,000/π
(6000/π)*.74 = 1414 spheres can be fit "in" each box.
