Human hamster ball

If the Zorb's outer sphere has a circumference of $9$ meters, then $2\pi R = 9 \Rightarrow R = \frac{9}{2\pi}.$ This same sphere is modeled as:

$x^2 + y^2 + z^2 = (\frac{9}{2\pi})^2$ (i).

If the inner cube has side length $s$ and its center is coincident with the outer sphere, then applying the planes $z = \pm \frac{s}{2}$ into (i) yields the circular plane section:

$x^2 + y^2 = (\frac{9}{2\pi})^2 - (\frac{s}{2})^2$ (ii)

and solving for the cube's side length can be determined by the following Pythagorean relationship:

$(\frac{s}{2})^2 + (\frac{s}{\sqrt{2}})^2 = (\frac{9}{2\pi})^2 \Rightarrow \frac{3s^2}{4} = (\frac{9}{2\pi})^2 \Rightarrow s = \frac{9}{\pi \sqrt{3}}$ (iii)

The inner-sphere has radius $r = \frac{s}{2}$ as it is tangent with all six sides of the inner cube. The volume computes to:

$V = \frac{4\pi}{3} r^3 = \frac{4\pi}{3} \cdot (\frac{9}{2\pi \sqrt{3}})^3 = \boxed{\frac{27 \sqrt{3}}{2 \pi^{2}}} .$

Hence, our desired final sum is $27 + 3 + 2 = \boxed{32}.$

$Diameter ~of~ the ~outer ~sphere~=\dfrac 9 \pi=~Diagonal~ of~ the~ cube.\\ Side~ of~ the~ cube~=\dfrac 1{ \sqrt3}*\dfrac 9 \pi.\\ Volume~ of~ inner~ sphere~=\dfrac 4 3*\pi*\left ( \dfrac 1 2 *\dfrac 1 {\sqrt3}*\dfrac 9 \pi. \right)^3\\ =\dfrac 4 3*\pi*\dfrac 1 8 *\dfrac 1{3\sqrt3}*\dfrac {9^3}{\pi^3}\\ = \dfrac 1 2 * \dfrac 1{9\sqrt3} *\dfrac {9*81}{{\pi}^2} \\ = \dfrac 1 2 * \dfrac {\sqrt3} 3 *\dfrac {81}{{\pi}^2}\\ = \dfrac 1 2 * \sqrt3 *\dfrac {27}{{\pi}^2}\\ = \dfrac {\sqrt B *A}{C*\pi^C} \\ A+B+C=27+3+2=\Large \color{#D61F06}{32}$