= Solution
Use the large-<radius> condensational-growth approximation from part (a), with fixed ambient <atmospheric supersaturation> and transport resistances. It gives
$$
\frac{d(r^2)}{dt}=\frac{2(S-1)}{\rho(F_k^l+F_d^l)},\qquad
\boxed{t=\frac{\rho(F_k^l+F_d^l)(r^2-r_0^2)}{2(S-1)}.}
$$
Taking $\rho=10^3\,\mathrm{kg\,m^{-3}}$, $S-1=0.01$, and negligible initial $r_0^2$ gives
$$
\boxed{t_a\simeq202.5\,\mathrm s\simeq3.4\,\mathrm{min},\qquad
t_b\simeq9.0\times10^5\,\mathrm s\simeq10.4\,\mathrm{days}.}
$$
The squared-<radius> law makes <condensation> relatively fast for reaching <cloud>-droplet sizes but far too slow for producing millimetre rain within a typical evolving <cloud>. Collision–coalescence, or ice growth followed by aggregation/riming and melting, bypasses this bottleneck. In reality <atmospheric supersaturation> is shared by many particles and changes with ascent and vapour depletion, so even the constant-$1\%$ estimate is idealized.
These are growth times \i[after activation], with negligible activation delay. The statement $r_0\ll1\,\mu\mathrm m$ alone does not ensure that a seed grows: for pure <water> with $B=0$, a seed with $r_0<A/(S-1)$ has a negative growth rate and shrinks. For soluble seeds the Köhler barrier depends on $A,B$ and the environmental <atmospheric supersaturation>. The literal full-model time is
$$
t=\rho(F_k^l+F_d^l)\int_{r_0}^{r}\frac{r'\,dr'}{(S-1)-A/r'+B/r'^3},
$$
which cannot be assigned a unique initial-stage numerical value without those microphysical parameters. The two boxed estimates are the intended continuum <condensation> scales for activated droplets, rather than a claim that every arbitrarily small seed reaches those sizes.
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