I think this is right...
I take it that if the hole is 6 inches deep and goes all the way through, then there are a limited set of solutions- one for each sphere. As the sphere gets larger, the edges of he whole move nearer (in terms of angle not necessarily size) to each other.
If we take a calculus based approach here, a sphere of radius r is made up of a series of infinitessimal disks, but when the cylindrical hole is drilled, it becomes a series of infinitessimal annuli. moving upwards from the "equator" with h increasing , the radius, y, of any one disk is given by
[math]y=(r^2 -h^2)^\frac{1}{2} [/math]
Taking a cylindrical hole of radius x out with the cross section of the cylinder to what I previously called the "equator", the width on an annulus w is,
[math]w=(r^2 -h^2)^\frac{1}{2} -x [/math]
And when the value of h is equal to 3 (i.e. the limit of a 6 inch hole, half being each side of the equator) the width of the annulus is 0, thus,
[math]0=(r^2 -3^2)^\frac{1}{2} -x [/math]
Which gives
[math]x=(r^2 -9)^\frac{1}{2} [/math]
And the volume of an annular element is thus given by
[math]\frac{dV}{dh}=\pi((r^2 -h^2)^\frac{1}{2})^2-\pi((r^2 -9)^\frac{1}{2})^2 [/math]
[math]\frac{dV}{dh}=\pi(r^2 -h^2-r^2 +9) [/math]
[math]\frac{dV}{dh}=\pi(9 -h^2) [/math]
Summing for the hole depth of 6 inches,
[math]V=\pi\int_{-3}^{3}(9 -h^2)dh [/math]
[math]V=\pi[9h-\frac{h^3}{3}]^3_-3[/math]
[math]V=\pi(27-9)-\pi(-27+9)=36\pi inches^3[/math]