 Solution

Let $r_s^2=N^{-1}\int r^{2}n(r)dr$. The effective droplet radius and shape factor give $r_e=r_v^3/r_s^2$ and $k^{1/3}=r_v/r_e$, so $r_s^2=k^{1/3}{r}_v^2$. If the extinction efficiency is approximately independent of droplet size over the distribution,For visible wavelengths and droplets large compared with the wavelength, geometrical optics makes $Q_{\mathrm{ext}}\simeq 2$ a useful approximation. Retaining it as the specified constant, and taking $N,k,c_w$ constant over the cloud,The $3/5$ comes from integrating $h^{2/3}$; replacing the whole cloud by its cloud-top extinction would incorrectly omit it. The effective-radius relation treats the distribution shape through $k$, without assuming all droplets have the same size. Liquid phase, negligible vertical variation in that shape factor, and negligible departures from the adiabatic water profile are essential assumptions.