[tex]\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad
\begin{cases}
B=\textit{area of the base}\\
h=height
\end{cases}[/tex]
now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12
now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8
now the last one on the far-right
has a height of 8, the Base is a Hexagon, with sides of 6
[tex]\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=6
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)[/tex]
