Explanation
The volume of a pyramid with a square base \(\frac{1}{3}\mathrm{e}^{2 } \times \text {height}\)
Given a volume of 125 and height of 15, here is how you can solve for \(\mathrm{e}^{2}\):
\(125= \frac{1}{3}\mathrm{e}^{2 } \times 15\)
\(375=15 \mathrm{e}^{2}\)
\(25=e^{2}\)