Check the picture below.
so let's firstly find the radius of that circular cross-section, since it's also the radius of the sphere.
[tex]\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ \cline{1-1} A=224.32 \end{cases}\implies 224.32=\pi r^2\implies \cfrac{224.32}{\pi }=r^2 \\\\\\ \sqrt{\cfrac{224.32}{\pi }}=r\implies 8.45\approx r \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\qquad \qquad \implies V=\cfrac{4\pi (8.45)^3}{3}\implies V\approx 2527.31[/tex]
