Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can see that $\rm He^+$ is a hydrogen-like ion with $Z=2$, so its energy levels are given by
$$E_n=-Z^2\dfrac{13.6}{n^2}=-(2^2)\dfrac{13.6}{n^2}$$
$$E_n=- \dfrac{54.4}{n^2} $$
So, the first five energy levels are
$$E_1= \dfrac{-54.4}{1^2}=\color{red}{\bf -54.4}\;\rm eV $$
$$E_2= \dfrac{-54.4}{2^2}=\color{red}{\bf -13.6}\;\rm eV $$
$$E_3= \dfrac{-54.4}{3^2}=\color{red}{\bf -6.04}\;\rm eV $$
$$E_4= \dfrac{-54.4}{4^2}=\color{red}{\bf -3.4}\;\rm eV $$
$$E_5= \dfrac{-54.4}{5^2}=\color{red}{\bf -2.18}\;\rm eV $$
$$\color{blue}{\bf [b]}$$
The ionization energy is given by
$$\Delta E=E_\infty-E_1=0-(-54.4)=\color{red}{\bf 54.4}\;\rm eV $$
where the ionization is at $n=\infty$
$$\color{blue}{\bf [c]}$$
All the possible emission transitions from the $n = 4$ energy level are
$$\lambda_{4\rightarrow 3}, \;\lambda_{4\rightarrow 2},\;\lambda_{4\rightarrow 1}$$
$$\color{blue}{\bf [d]}$$
The magnitudes of these 3 transitions are given by
$$\lambda=\dfrac{hc}{\Delta E}$$
So,
$$\lambda_{4\rightarrow 3}=\dfrac{hc}{|E_3-E_4|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-6.04-(-3.4)|(1.6\times 10^{-19})}$$
$$\lambda_{4\rightarrow 3}=\color{red}{\bf 471}\;\rm nm$$
$$\lambda_{4\rightarrow 2}=\dfrac{hc}{|E_2-E_4|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-13.6-(-3.4)|(1.6\times 10^{-19})}$$
$$\lambda_{4\rightarrow 3}=\color{red}{\bf 122}\;\rm nm$$
$$\lambda_{4\rightarrow 1}=\dfrac{hc}{|E_1-E_4|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-54.4-(-3.4)|(1.6\times 10^{-19})}$$
$$\lambda_{4\rightarrow 3}=\color{red}{\bf 24.4}\;\rm nm$$
