I can't speak to the cactus specifically, but any sphere has a surface area of [tex]4\pi r^2[/tex] and a volume of [tex] \frac{4}{3} \pi r^3[/tex], so the area-to-volume ratio wold be
[tex] \frac{4\pi r^2}{ \frac{4}{3}\pi r^3 } = \frac{3(4\pi r^2)}{4(\pi r^3)}[/tex]
[tex]= \frac{3\pi r^2}{\pi r^3}\\\\
= \frac{3}{r} [/tex]
Where r is the radius of the sphere.
