Answer
${\bf 4.59 }\;\rm eV$
Work Step by Step
We know, from Figure 41.24, that the energy of the $ 3p $ sodium state is $ E_{3p} = 2.104 \, \text{eV} $.
We need to find the energy of the $ 5d $ state and we can do that by finding the energy of the emitted photon which is equal to the energy difference between $3p$ and $5s$.
Then we need to add the transition energy from $ 5d \to 3p $ to the energy of the $ 3p $ state to get the energy of the $5d$ state.
The transition energy from the $ 5d $ state to the $ 3p $ state is given by:
$$
E_{5d \to 3p} = \frac{hc}{\lambda}=\dfrac{(6.63\times 10^{-34}(3\times 10^8))}{(499\times 10^{-9})}$$
$$E_{5d \to 3p}=\bf 4\times 10^{-19}\;\rm J=\bf 2.49\;\rm eV
$$
Recalling that
$$E_{5d \to 3p} =E_{5d}-E_{3p}$$
So,
$$E_{5d}=E_{5d \to 3p}+E_{3p}$$
Plug the known;
$$E_{5d}=2.49+2.104 =\color{red}{\bf 4.59 }\;\rm eV$$
