Check the picture below.
so the volume will simply be the area of the hexagonal face times the height.
[tex]\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\stackrel{\qquad degrees}{\cot\left( \frac{180}{n} \right)}~~ \begin{cases} n=\stackrel{number~of}{sides}\\ s=\stackrel{length~of}{side}\\[-0.5em] \hrulefill\\ n=6\\ s=12 \end{cases}\implies A=\cfrac{1}{4}(6)(12)^2\cot\left( \frac{180}{6} \right) \\\\\\ A=216\cot(30^o)\implies A=216\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the hexagon}}{(216\sqrt{3})}~~\stackrel{height}{(10)}\implies 2160\sqrt{3}~~\approx ~~3741.2~cm^3[/tex]
