Sphere inscribed in a hexagonal truncated pyramid.

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

Let $r^{*}$ be the lower base of the hexagonal truncated pyramid and $r^{**}$ be the upper base of the hexagonal truncated pyramid, where $r^{*} > r^{**}$ .

From the diagram above the altitudes of the upper and lower bases are $h^{*} = \dfrac{\sqrt{3}}{2} r^{*}$ and $h^{**} = \dfrac{\sqrt{3}}{2} r^{**}$ respectively.

$h^{*} = 10 - h^{**} = \dfrac{\sqrt{3}}{2} r^{*}$ and $10 - 2h^{**} = 10 -\sqrt{3}r^{**} \implies 64 + (10 - \sqrt{3}r^{**})^2 = 100 \implies$ $64 + 100 - 20\sqrt{3}r^{**} + 3({r^{**}})^2= 100 \implies 3({r^{**}})^2 - 20\sqrt{3}r^{**} + 64 = 0 \implies r^{**} = \dfrac{10 \pm 6}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}, \:\ \dfrac{16}{\sqrt{3}}$ .

Letting the smaller of the two bases $r^{**} = \dfrac{4}{\sqrt{3}} \implies h^{**} = \dfrac{\sqrt{3}}{2}r^{**} = \dfrac{\sqrt{3}}{2} * \dfrac{4}{\sqrt{3}} = 2 \implies$

$h^{*} = 10 - h^{**} = 8 = \dfrac{\sqrt{3}}{2}r^{*} \implies r^{*} = \dfrac{16}{\sqrt{3}}$ .

The volume of a $n$ -gonal truncated pyramid is $V = \dfrac{nh}{12}((r^{*})^2 + r^{*}r^{**} + (r^{**})^2) \cot(\dfrac{\pi}{n})$ For $n = 6 \implies V = \sqrt{3}R((r^{*})^2 + r^{*}r^{**} + (r^{**})^2) = \dfrac{4\sqrt{3}(256 + 64 + 16)}{3} = \boxed{448\sqrt{3}} \approx \boxed{775.95876}$