Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
For the absorption spectrum, we know that the energy absorbed by the atom is given by
$$E_{\rm photon}=\Delta E_{\rm atom}=\dfrac{hc}{\lambda}$$
We are given 3 wavelengths, and we need to find the energy absorbed by the atom in each case in eV.
So,
$$ ( \Delta E_{\rm atom})_1=\dfrac{hc}{e\lambda_1}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})(200\times 10^{-9})}=\bf 6.22\;\rm eV$$
$$ ( \Delta E_{\rm atom})_2=\dfrac{hc}{e\lambda_2}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})(300\times 10^{-9})}=\bf 4.14\;\rm eV$$
$$ ( \Delta E_{\rm atom})_3=\dfrac{hc}{e\lambda_3}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})(500\times 10^{-9})}=\bf 2.49\;\rm eV$$
Now we can draw the atom’s energy-level diagram since we can assume that these 3 energies are for $n=2,3,4$.
See the figure below.
$$\color{blue}{\bf [b]}$$
The emitted wavelengths, as we see from the figure below, are given by
$$\lambda_{4\rightarrow3 }=\dfrac{hc}{\Delta E_{4\rightarrow 3}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(6.22-4.14)(1.6\times 10^{-19})}=\color{red}{\bf 598}\;\rm nm$$
$$\lambda_{4\rightarrow2 }=\dfrac{hc}{\Delta E_{4\rightarrow 2}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(6.22-2.49)(1.6\times 10^{-19})}=\color{red}{\bf 333}\;\rm nm$$
$$\lambda_{4\rightarrow1 }=\dfrac{hc}{\Delta E_{4\rightarrow 1}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(6.22-0)(1.6\times 10^{-19})}=\color{red}{\bf 200}\;\rm nm$$
$$\lambda_{3\rightarrow2 }=\dfrac{hc}{\Delta E_{3\rightarrow 2}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(4.14-2.49)(1.6\times 10^{-19})}=\color{red}{\bf 753}\;\rm nm$$
$$\lambda_{3\rightarrow1 }=\dfrac{hc}{\Delta E_{3\rightarrow 1}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(4.14-0)(1.6\times 10^{-19})}=\color{red}{\bf 300}\;\rm nm$$
$$\lambda_{2\rightarrow1 }=\dfrac{hc}{\Delta E_{2\rightarrow 1}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(2.49-0)(1.6\times 10^{-19})}=\color{red}{\bf 499}\;\rm nm$$
