Answer
$\rm 10.3\;nm,\;7.62\;nm,\;6.80\;nm$
Work Step by Step
We can see that $\rm O^{+7}$ is a hydrogen-like ion with $Z=8$ since it has one electron rotating around its nucleus.
So its energy levels are given by
$$E_n=-Z^2\dfrac{13.6}{n^2}=-(8^2)\dfrac{13.6}{n^2}$$
$$E_n=- \dfrac{870.4}{n^2}\tag 1 $$
We need to find the energy levels up to the fifth level since the author asks for the transitions up to the fifth level.
So, the first five energy levels are
$$E_1= \dfrac{-870.4}{1^2}=\color{blue}{\bf -870.4}\;\rm eV $$
$$E_2= \dfrac{-870.4}{2^2}=\color{blue}{\bf -217.6}\;\rm eV $$
$$E_3= \dfrac{-870.4}{3^2}=\color{blue}{\bf -96.71}\;\rm eV $$
$$E_4= \dfrac{-870.4}{4^2}=\color{blue}{\bf -54.4}\;\rm eV $$
$$E_5= \dfrac{-870.4}{5^2}=\color{blue}{\bf -34.82}\;\rm eV $$
Now we need to find
$$\lambda_{3\rightarrow 2}, \;\lambda_{4\rightarrow 2},\;\lambda_{5\rightarrow 2}$$
Recalling that
$$\lambda=\dfrac{hc}{\Delta E}$$
So,
$$\lambda_{3\rightarrow 2}=\dfrac{hc}{|E_2-E_3|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-96.71-(-217.6)|(1.6\times 10^{-19})}$$
$$\lambda_{3\rightarrow 2}=\color{red}{\bf 10.3}\;\rm nm$$
$$\lambda_{4\rightarrow 2}=\dfrac{hc}{|E_2-E_4|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-217.6-(-54.4)|(1.6\times 10^{-19})}$$
$$\lambda_{4\rightarrow 3}=\color{red}{\bf 7.62}\;\rm nm$$
$$\lambda_{5\rightarrow 2}=\dfrac{hc}{|E_2-E_5|}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{|-217.6-(-34.82)|(1.6\times 10^{-19})}$$
$$\lambda_{5\rightarrow 3}=\color{red}{\bf 6.80}\;\rm nm$$
We know that the visible light range is 400 nm $\rightarrow$ 700 nm.
All these 3 wavelengths are less than 400 nm, so they all are ultraviolet.
