Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
A sodium atom in the $ 6s $ state can undergo downward transitions to lower energy states.
For typical sodium energy levels, assuming similar to the hydrogen-like levels, the possible downward transitions are:
$$\boxed{ 6s \to 3p\quad, \quad 6s \to 4p \quad, \quad 6s \to 5p }$$
$$\color{blue}{\bf [b]}$$
We know that
$$
E_{\rm photon} =\Delta E= \frac{hc}{\lambda}
$$
Where $ \Delta E $ is the energy difference between the two states.
So,
$$
\lambda = \frac{hc}{ \Delta E}\tag 1
$$
According to the data provided in the mentioned figure, we can see that
$$\Delta E_{\;6s\to 3p}=4.51-2.104=\bf 2.406 \;\rm eV$$
$$\Delta E_{\;6s\to 4p}=4.51-3.75=\bf 0.76\;\rm eV$$
$$\Delta E_{\;6s\to 5p}=4.51-4.34=\bf 0.17\;\rm eV$$
Using these energy differences and (1) to find the emitted wavelengths.
$$\lambda_{\;6s\to 3p}= \frac{hc}{ \Delta E_{\;6s\to 3p}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(2.406 )(1.6\times 10^{-19})}$$
$$\lambda_{\;6s\to 3p}=\color{red}{\bf 517}\;\rm nm$$
$$\lambda_{\;6s\to 4p}= \frac{hc}{ \Delta E_{\;6s\to 4p}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(0.76)(1.6\times 10^{-19})}$$
$$\lambda_{\;6s\to 4p}=\color{red}{\bf 1636}\;\rm nm$$
$$\lambda_{\;6s\to 5p}= \frac{hc}{ \Delta E_{\;6s\to 5p}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(0.17)(1.6\times 10^{-19})}$$
$$\lambda_{\;6s\to 5p}=\color{red}{\bf 7313}\;\rm nm$$
