The longest  wavelength  doublet  absorption  transition is observed  at 589 and  589.6  nm.  Energy difference  between two  excited  states is

step I : Wavelength ($\mathrm{\lambda }$) = 589 nm $=589\times {10}^{-9}\mathrm{m}$
Frequency $\left(\mathrm{v}\right)=\frac{\mathrm{c}}{\mathrm{\lambda }}=\frac{3\times {10}^8}{589\times {10}^{-9}}$
$=5.093\times {10}^{14}\text{ cycles per sec }$
step II : Wavelenght $\left(\mathrm{\lambda }\right)=589.6\mathrm{nm}=589.6\times {10}^{-9}\mathrm{m}$
$\therefore \mathrm{v}=\frac{\mathrm{c}}{\mathrm{\lambda }}=\frac{3\times {10}^8}{589.6\times {10}^{-9}}=5.088\times {10}^{14}$ cycles per sec
Energy difference between two excited states,
$\begin{array}{l}\mathrm{\Delta E}=6.626\times {10}^{-34}(5.093-5.088){10}^{14}\\ =6.626\times {10}^{-34}\times 5\times {10}^{-3}\times {10}^{14}=3.31\times {10}^{-22}\mathrm{J}\end{array}$