check the picture below.
thus, the radius of each is 4 inches, and the heights are 9 and "h" respectively.
the volume of the figure is the volume of the cone plus the volume of the cylinder then, thus
[tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\qquad \qquad \implies V=\cfrac{\pi 4^2\cdot 9}{3}\implies V=48\pi
\\\\\\
\textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad \qquad \implies V=\pi 4^2\cdot h\implies V=16\pi h\\\\
-------------------------------\\\\
\stackrel{\textit{cone's volume}}{48\pi }+\stackrel{\textit{cylinder's volume}}{16\pi h}=653.12\implies 16\pi h=653.12-48\pi
\\\\\\
h=\cfrac{653.12-48\pi }{16\pi }[/tex]
