Answer
$4.29\;\mu m$
Work Step by Step
Here, the carbon dioxide molecules are excited from energy level $E_0$ to energy level $E_2$ in the lasing process.
Thus, the energy associated with the sunlight that excites the molecules in the lasing action is:
$ \Delta E=E_2-E_0=(0.289-0)\;eV=0.289\;eV$
Therefore, the wavelength of sunlight that excites the molecules in the lasing action is:
$\lambda=\frac{ch}{\Delta E}$
or, $\lambda=\frac{1240\;eV-nm}{\Delta E \;(eV)}$
or, $\lambda=\frac{1240\;eV-nm}{0.289 \;eV}$
or, $\lambda=4.29\times10^{3}\;nm$
or, $\lambda=4.29\;\mu m$
