Inscribed three dimensional objects.

$V_{p} = \dfrac{1}{3} x^2 H$ and $V_{c} = \pi r^2 H$

Cutting the right circular cylinder of height $h$ and radius $r$ inscribed in the square pyramid by a single plane oriented parallel to the cylinder's axis of symmetry we obtain the proportion:

$\dfrac{x}{2H} = \dfrac{x - 2r}{2h} \implies r = \dfrac{x}{2H} (H - h) \implies V_{c}(h) = (\dfrac{\pi x^2}{4 H^2}) (h^3 - 2H h^2 + H^2 H) \implies$

$\dfrac{V_{c}}{dh} = (\dfrac{\pi x^2}{4 H^2}) (3 h^2 - 4H h + h^2) = 0 \implies h = H, \dfrac{H}{3}$

$h \neq H \implies h = \dfrac{H}{3}$

$\dfrac{d^2 V_{c}}{dh^2}|_{(h = \dfrac{H}{3})} = \dfrac{-\pi x^2}{2H} < 0 \implies$ $h = \dfrac{H}{3}$ maximizes $V_{c}(h)$

$h = \dfrac{H}{3} \implies r = \dfrac{x}{3} \implies V_{c} = \dfrac{\pi}{9} V_{p}$ .

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Sorry, but the above picture was the only picture that I could find. Assume the base is a square base and assume four faces.

Using the above picture:

Let $x$ be the side of the square base and $AH$ be the height $h$ of the pyramid. Let $OA$ and $OC$ be the radius $R$ of the sphere, so $OH$ is $h - R$ and half the diagonal $CH$ of the square is $\dfrac{x}{\sqrt{2}}$ . The volume of the sphere $V_{s} = \dfrac{4}{3} \pi R^3$ and the volume of the square pyramid $V_{p} = \dfrac{1}{3} x^2 h$ .

For right triangle $COH$ we have:

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$R^2 = \dfrac{x^2}{2} + (h - R)^2 = \dfrac{x^2}{2} + h^2 - 2hR + R^2 \implies x^2 + 2h^2 - 4hR = 0 \implies x^2 = 4hR - 2h^2 \implies$ $$

$V_{p}(h) = \dfrac{1}{3} (4h^2R - 2h^3) \implies \dfrac{dV_{p}}{dh} = \dfrac{2h}{3}(4R - 3h) = 0$ and $h \neq 0 \implies h = \dfrac{4R}{3}.$ $$

$h = \dfrac{4R}{3}$ maximizes $V_{p}(h)$ since $\dfrac{d^2V_{p}}{dh^2}|_{(h = \dfrac{4R}{3})} =\dfrac{-8R}{3} < 0$
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$h = \dfrac{4R}{3} \implies x^2 = \dfrac{16 R^2}{9} \implies x = \dfrac{4R}{3} = h \implies V_{p} = \dfrac{16}{27\pi} V_{s}$

$V_{c} = \dfrac{\pi}{9} V_{p} \implies V_{c} = \dfrac{16}{243} V_{s}$

Letting $V_{s} = 243 \implies V_{c} = \boxed{16}$ .