Answer
$16\pi$
Work Step by Step
Given $S(r)=4\pi r^2S$, the rate of change of the surface area with respect to the radius at $r=2$ is given by:
$$S'=\lim_{h\to 0}\frac{S(2+h)-S(2)}{h}=\lim_{h\to 0}\frac{4\pi(2+h)^2-4\pi2^2}{h}=\lim_{h\to 0}\frac{4\pi(4h+h^2)}{h}=\lim_{h\to 0}4\pi(4+h)=16\pi$$