if the ball has a diameter of 8.5, then its radius is half that, or 4.25.
[tex]\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=4.25 \end{cases}\implies V=\cfrac{4\pi (4.25)^3}{3}\implies V=\cfrac{4913\pi }{48} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill V\approx 321.56~\hfill[/tex]
