Find the volume of a sphere with a great circle area of 201.06 square inches

assuming you meant a "circular cross-section" of 201.06 in².

so, we know if we cut the sphere like a cantaloupe, in half, the circular inner-part of the cantaloupe will have an area of 201.6 in².

keeping in mind that, the radius of that circular section, is the same radius of the sphere, what is it anyway?

[tex]\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ A=201.06 \end{cases}\implies 201.06=\pi r^2 \\\\\\ \cfrac{201.06}{\pi }=r^2\implies \sqrt{\cfrac{201.06}{\pi }}=r\\\\ -------------------------------[/tex]

[tex]\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\qquad \qquad \implies V=\cfrac{4\pi \left( \sqrt{\frac{201.06}{\pi }} \right)^3}{3}\implies V=\cfrac{4\pi \left( \frac{\sqrt{201.06^3}}{\sqrt{\pi ^3}} \right)}{3} \\\\\\ V=\cfrac{4\pi \cdot \frac{\sqrt{201.06^3}}{\pi \sqrt{\pi }}}{3}\implies V=\cfrac{\frac{4\sqrt{201.06^3}}{\sqrt{\pi }}}{3}\implies V=\cfrac{4\sqrt{201.06^3}}{3\sqrt{\pi }}[/tex]