Let radius of sphere $A$ = $r_a$
Let radius of sphere $B$ = $r_b$
Surface area of sphere = $4\pi r^2$
Volume of sphere = $\frac{4}{3} \pi r^3$
Let surface area of sphere $A = x$
Let surface area of sphere $B = x + 3x = 4x$ [Since surface area of B is 300% higher than A]
$\implies \large{\frac{\text{Surface area of sphere A}}{\text{Surface area of sphere B}} = \frac{4 \pi r_a^2}{4 \pi r_b ^2}}$
$\implies \large{\frac{x}{4x}} = (\frac{r_a}{r_b})^2$
$\implies \large{\frac{r_a}{r_b} = \frac{1}{2}}$
Let volume of sphere $B = y$
Let volume of sphere $A = y – \frac{k}{100}y = \frac{(100-k)y}{100}$
$\implies \large{\frac{\text{Volume of sphere A}}{\text{Volume of sphere B}} = \frac{\frac{4}{3} \pi r_a^3}{\frac{4}{3} \pi r_b^3}}$
$\implies \large{\frac{\frac{(100-k)y}{100}}{y} = (\frac{r_a}{r_b}})^3$
$\implies \large{\frac{100-k}{100} = \frac{1}{8}}$
$\implies 800-8k = 100 \implies 8k =700 \implies k = 87.5$
