= Solution
For the second transition, $r=3$ and the much longer lower-level lifetime changes the sign of the available <population inversion>:
$$
A^{-1}-r\tau_1=\frac1{3\times10^6}-3(3\times10^{-7})
=-5.667\times10^{-7}\ \mathrm s,
\qquad D_0=-2.833\times10^{13}\ \mathrm{cm^{-3}}.
$$
If its <emission cross section> equals the first transition's value, the estimate is
$$
\boxed{\alpha_0\simeq-0.1133\ \mathrm{cm^{-1}}=-11.33\ \mathrm{m^{-1}}.}
$$
The word “comparable” does not specify exact equality, so the more general answer is $\alpha_0=-(0.1133\ \mathrm{cm^{-1}})[\sigma_2/(4\times10^{-15}\ \mathrm{cm^2})]$. The quoted numerical value is an explicitly equal-cross-section estimate, while the negative sign is robust for any positive comparable <emission cross section>.
In this case $rA\tau_1=2.7>1$. At fixed pump, $N_1=1.50\times10^{13}$ cm$^{-3}$, so its degeneracy-weighted contribution exceeds the upper population. The medium therefore absorbs a weak beam at this transition. Increasing the radiation saturates the <photon absorption> magnitude toward zero according to <3/c>; it does not convert this fixed-pump model into an <optical amplifier> with positive gain. Simply inserting the magnitude $|\alpha_0|$ into the propagation law for positive <optical gain> would discard the physically important population balance.
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