\(\displaystyle{V}={\left(\frac{{1}}{{3}}\right)}{B}{h}\)

\(\displaystyle{120}={\left(\frac{{1}}{{3}}\right)}{\left({36}\right)}{h}\)

\(120=12h\)

\(10=h\)

From part (a), the volume of the clay is \(120 \ln^3\). Let the volume of the pyramid be \(V=120 \ln^3\). If the square base has dimensiont of 6 in, by 6 in, then it has an area of \(B=6^2=36 \ln^2\). Substitute in the volume and area of the base into the formula for the volume of a pyramid. Then solve for h.