For small Reynold number, the viscous force on the small sphere of radius #r# moving with a velocity #v# in a fluid of dynamic viscosity #\mu# is given by Stoke's law as
#F_v=6\pi\mu r v#
For terminal or critical velocity #v_c# to achieve, the viscous force on the small sphere (density #\rho_s#) must be equal to the net weight of body in the fluid of density #\rho_f# hence we have
#F_v=mg-\text{Buoyant force}#
#6\pi\mur v_c={4\pi}/3 r^3(\rho_s-\rho_f)g#
#v_c=\frac{2}{9}\frac{(\rho_s-\rho_f)gr^2}{mu}#
#v_c\propr^2#
the above relation shows that the terminal velocity #v_c# of the sphere is directly proportional to the square of radius #r# keeping other parameters constant
Now, 8 identical water drops each of radius #r_1=2\ mm# collapse to form a single big drop of radius #r_2#
Now, by the conservation of volume, volume of big drop of water (radius #r_2#) must be equal to the volume of 8 identical water drops #(r_1=2mm )# then we have
#\frac{4\pi}{3}r_2^2=8\times \frac{4\pi}{3}r_1^2#
#r_2=2r_1#
#r_2=2(2)=4\ mm#
From, terminal velocity relation (derived above), we know
#v_c\propr^2#
#\frac{(v_c)_2}{(v_c)1}=\frac{r_2^2}{r_1^2}#
#\frac{(v_c)_2}{8}=\frac{4^2}{2^2}#
#\frac{(v_c)_2}{8}=4#
#(v_c)_2=32\ \text{cm/sec}#
