Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. Determine the value of k.

Let the radius of sphere A be 'r' and sphere B be 'R'
Surface area of A = $$4\pi r^2$$
Surface area of B = $$4\pi R^2$$ = (1+300%) of A = 4 * $$4\pi r^2$$
R=2r
Volume of A = $$\frac{4}{3}\pi r^3$$ 
Volume of B = $$\frac{4}{3}\pi R^3\ =\ \ \frac{\ 4}{3}\pi\ \left(2r\right)^3\ =\ \ \frac{\ 32}{3}\pi\ r^3$$
Required percent = $$\ \frac{\ \left(\frac{32}{3}\pi\ r^3\ -\ \frac{4}{3}\pi\ r^3\right)}{\ \frac{32}{3}\pi\ r^3}\times\ 100\ =87.5\ \%\ $$
Option C is the answer.