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]