1. **State the problem:** We have a sphere of radius $R=18$ cm and a right circular cone inscribed inside it with height $h$ and base radius $r$. We want to find the maximum volume of this cone in terms of $\pi$. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ 3. **Relate $r$, $h$, and $R$ using the geometry:** The cone fits inside the sphere, so the radius $r$ and height $h$ satisfy the Pythagorean relation from the right triangle formed: $$r^2 + (R - h)^2 = R^2$$ 4. **Express $r^2$ in terms of $h$ and $R$:** $$r^2 = R^2 - (R - h)^2 = R^2 - (R^2 - 2Rh + h^2) = 2Rh - h^2$$ 5. **Substitute $r^2$ into volume formula:** $$V = \frac{1}{3} \pi (2Rh - h^2) h = \frac{1}{3} \pi (2Rh^2 - h^3)$$ 6. **Simplify volume expression:** $$V = \frac{\pi}{3} (2R h^2 - h^3)$$ 7. **Find $h$ that maximizes $V$ by differentiation:** $$\frac{dV}{dh} = \frac{\pi}{3} (4R h - 3 h^2) = 0$$ 8. **Solve for $h$:** $$4R h - 3 h^2 = 0 \implies h(4R - 3h) = 0$$ Ignoring $h=0$, we get: $$4R - 3h = 0 \implies h = \frac{4R}{3}$$ 9. **Calculate $r^2$ at this $h$:** $$r^2 = 2R h - h^2 = 2R \cdot \frac{4R}{3} - \left(\frac{4R}{3}\right)^2 = \frac{8R^2}{3} - \frac{16R^2}{9} = \frac{24R^2 - 16R^2}{9} = \frac{8R^2}{9}$$ 10. **Calculate maximum volume $V_{max}$:** $$V_{max} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \cdot \frac{8R^2}{9} \cdot \frac{4R}{3} = \frac{1}{3} \pi \cdot \frac{32 R^3}{27} = \frac{32 \pi R^3}{81}$$ 11. **Substitute $R=18$ cm:** $$V_{max} = \frac{32 \pi (18)^3}{81} = \frac{32 \pi \cdot 5832}{81} = 32 \pi \cdot 72 = 2304 \pi$$ **Final answer:** $$\boxed{2304 \pi}$$ This corresponds to option A.