= Solution
Keep the given <emission cross section> in square centimetres so that the density remains per cubic centimetre. For the first transition, $r=5/3$ and
$$
A^{-1}-r\tau_1=\frac1{3\times10^6}-\frac53(2\times10^{-8})
=3.00\times10^{-7}\ \mathrm s.
$$
Thus
$$
D_0=(5\times10^{19})(3.00\times10^{-7})=1.50\times10^{13}\ \mathrm{cm^{-3}},
\qquad\boxed{\alpha_0=(4\times10^{-15})D_0=0.0600\ \mathrm{cm^{-1}}=6.00\ \mathrm{m^{-1}}.}
$$
The separate populations are $N_2^{(0)}=1.667\times10^{13}$ cm$^{-3}$ and $N_1^{(0)}=1.00\times10^{12}$ cm$^{-3}$. Subtracting $5N_1/3$, rather than $N_1$ alone, reproduces the <population inversion> used above.
Here $rA\tau_1=0.10<1$, so upper-state pumping and sufficiently rapid lower-level removal produce positive small-signal <optical gain>. This coefficient is an <intensity> <optical gain>; a weak field amplitude would grow with half this exponent. A positive single-pass coefficient also does not by itself establish <laser oscillation> without a cavity-loss budget. No transition <wavelength> was supplied for this numerical example, so a numerical <saturation intensity> is not needed or invented to calculate $\alpha_0$.
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