check the picture below.
so, now, we simply get the volume of the cylinder, and then volume of the 3 balls, and subtract the volume of the balls from the cylinders, effectively making a hole in the volume of the cylinder, and what's leftover, is the empty area.
[tex]\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h~
\begin{cases}
r=3.4\\
h=20.4
\end{cases}\implies V=\pi \cdot 3.4^2\cdot 20.4\implies V=235.824\pi \\\\
-------------------------------\\\\
\textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\qquad r=3.4\implies V=\cfrac{4\cdot \pi \cdot 3.4^3}{3}\implies V=\cfrac{157.216\pi }{3}[/tex]
[tex]\bf \textit{volume of the \underline{3 balls}}\quad 3\cdot \cfrac{157.216\pi }{3}\implies 157.216\pi \\\\
-------------------------------\\\\
\stackrel{\textit{volume of the cylinder}}{235.824\pi }-\stackrel{\textit{volume of the balls}}{157.216\pi }\implies 78.608\pi [/tex]
