8. The maximum volume of a cone inscribed in a sphere compared to the volume of the sphere is $\qquad$

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Consider the cone $S H$ inscribed in the sphere $O$, with $S H=h$,
$$
\begin{array}{l}
O S=R, H B=r \Rightarrow(h-R)^{2}+r^{2}=R^{2} \\
\Rightarrow h^{2}-2 h R+r^{2}=0 \Rightarrow r^{2}=h(2 R-h) .
\end{array}
$$

Then $V_{S H}=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi h^{2}(2 R-h)$
$\leqslant \frac{\pi}{6}\left(\frac{h+h+4 R-2 h}{3}\right)^{3}=\frac{32}{81} \pi R^{3}$, equality holds when $h=4 R-2 h \Rightarrow h=\frac{4 R}{3}$. Therefore, the required volume ratio is $\frac{32}{81} \times \frac{3}{4}=\frac{8}{27}$.