$It is given that the volume of the raindrop initially is $10^{-6}m^3$, so we may solve for the initial radius: $\\10^{-6} = \frac43\pi r^3 \implies r = \sqrt[3]{\frac{3\times10^{-6}}{4\pi}}\\$Using part i, we know that $ \frac{dr}{dt} = -10^{-4}$, so $ r = -10^{-4}t + C $ for some constant...