Sphere Inscribed In Frustum.

Cut the frustum and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the two bases.

Ignore the numbers in the diagram above.

Let the slant height $CB = 10$ and the radius of the circle $OG = OE = OF = 3$

Let $EC = r^{*}$ , since $\triangle{EOC} \cong \triangle{COF} \implies CF = r^{*}$ and $FB = 10 - r^{*}$ . Since $\triangle{FOB} \cong \triangle{BOG} \implies BG = 10 - r^{*}$ . From $C$ draw a perpendicular to base $AB$ at point $I$ , then $\triangle{CIB}$ is a right triangle, where $CI = 6, IB = 10 - 2r^{*}$ and $CB = 10 \implies (10 - 2r^{*})^2 + 36 = 100 \implies {r^{*}}^{2} - 10r^{*} + 9 = 0 \implies r^{*} = 9,1$ . Since the radius of the smaller base $r^{*} < R = 10 - r^{*}$ we choose $r^{*} = 1 \implies R = 10 - r^{*} = 9 \implies$ the volume of the frustum $V = \dfrac{\pi}{3} h(R^2 + r^{*}R + {r^{*}}^{2}) = \dfrac{\pi}{3} (2 * 3) (81 + 9 + 1) = \boxed{182\pi} = \boxed{571.76986295}$ to eight decimal places.