Answer
$1.146\times10^{6}\;W$
Work Step by Step
The Cr ions are in the first excited state: $N_1=0.6\times4\times10^{19}$
The Cr ions are in the first excited state: $N_0=0.4\times4\times10^{19}$
Total energy of the of the photons emitted in per pulse of the laser:
$E=(N_1-N_0)E_p$
where, $E_p$ is the energy of a photon
$E_p=\frac{ch}{\lambda}=\frac{3\times10^{8}\times6.63\times10^{-34}}{694\times10^{-9}}\;J=2.866\times10^{-19}\;J$
Therefore, the average power emitted during the pulse is
$P=\frac{E}{t}=\frac{(N_1-N_0)E_p}{t}$
or, $P=\frac{(0.6-0.4)\times4\times10^{19}\times2.866\times10^{-19}}{2\times10^{-6}}\;W=1.146\times10^{6}\;W$
