Step 1: Think in terms of the final equalized population.
Stimulated absorption and stimulated emission use the same rate constant here, so the process can only keep pushing atoms from the upper to the lower level while $N_2 > N_1$; it must stop exactly when the two populations become equal, since beyond that point absorption and emission balance and cancel. So instead of tracking the difference, look directly at the common final population both levels settle to.
Step 2: Find the final common population.
The total number of atoms, $N_1+N_2 = 5\times10^{20}+8\times10^{20}=13\times10^{20}$, stays fixed since only redistribution between the two levels happens (no atoms enter or leave). When the populations equalize, each level ends up with half this total:
\[ N_f = \frac{13\times10^{20}}{2} = 6.5\times10^{20} \]
Step 3: Find how many atoms moved down.
The upper level goes from $8\times10^{20}$ down to $6.5\times10^{20}$, so the number of atoms that made the downward transition is
\[ \Delta N = 8\times10^{20} - 6.5\times10^{20} = 1.5\times10^{20} \]
This matches checking the lower level too: it rises from $5\times10^{20}$ to $6.5\times10^{20}$, a gain of $1.5\times10^{20}$, consistent with $\Delta N$.
Step 4: Convert to energy in joules.
Each transferred atom releases one photon carrying the level gap of 2.2 eV, so
\[ E = \Delta N \times 2.2\text{ eV} \times e = (1.5\times10^{20})(2.2)(1.6\times10^{-19}) \]
\[ E = 3.3\times10^{20}\times1.6\times10^{-19} = 52.8\text{ J} \]
Final Answer:
Rounded to one decimal place,
\[ \boxed{E = 52.8\text{ J}} \]