Answer
$6.90\;\mu eV$
Work Step by Step
The energy difference $E_2-E_1$ for the given stimulated emission is equal to the energy of the emitted photon .
$E_2-E_1=hf$
or, $E_2-E_1=(4.14\times10^{-15}\;eV·s)\times(1666\;MHz)$
or, $E_2-E_1=(4.14\times10^{-15}\;eV·s)\times(1666\times10^{6}\;Hz)$
or, $E_2-E_1\approx6.90\times10^{-6}\;eV$
or, $E_2-E_1=6.90\;\mu eV$
Therefore, the energy difference $E_2-E_1$ for the given stimulated emission is $6.90\;\mu eV$
