Answer
See below
Work Step by Step
From part b, we found: $S=(4\pi)^{1/3}(3V)^{2/3}$
Substitute $2V$ for $V$:$S=(4\pi)^{1/3}(3(2V))^{2/3}\\=(4\pi)^{1/3}(3V)^{2/3}(2)^{2/3}$
Since $2^{2/3}\approx1.587$, the surface areas of the two water balloons have about $1.587$ times as much surface area.