Answer
$\rm 656.5\;nm,\;486.3\;nm,\; 434.2\;nm,\; 410.3\;nm $
Work Step by Step
We know that the wavelengths in the hydrogen spectrum are given by
$$\lambda_{n\rightarrow m}=\dfrac{\lambda_0}{\dfrac{1}{m^2}-\dfrac{1}{n^2}}$$
where $m=1,2,3,...$, and $n=m+1,m+2,m+3,...$
and $\lambda_0=91.18\;\rm nm$
$$\lambda_{n\rightarrow m}=\dfrac{91.18\;\rm nm}{\dfrac{1}{m^2}-\dfrac{1}{n^2}}\tag 1$$
For each $m$, the wavelengths range from $\lambda_{\rm max}$ at $n=m+1$ to $\lambda_{\rm min}$ at $n=\infty$.
$\bullet$ For $m=1$, using (1),
$$\lambda_{\rm max}=\dfrac{91.18}{\dfrac{1}{1^2}-\dfrac{1}{(1+1)^2}}=\color{blue}{\bf 122}\;\rm nm$$
$$\lambda_{\rm min}=\dfrac{91.18}{\dfrac{1}{1^2}-\dfrac{1}{(\infty)^2}}=\color{blue}{\bf 91.2}\;\rm nm$$
$\bullet$ For $m=2$, using (1),
$$\lambda_{\rm max}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{(2+1)^2}}=\color{red}{\bf 656}\;\rm nm$$
$$\lambda_{\rm min}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{(\infty)^2}}=\color{red}{\bf 365}\;\rm nm$$
$\bullet$ For $m=3$, using (1),
$$\lambda_{\rm max}=\dfrac{91.18}{\dfrac{1}{3^2}-\dfrac{1}{(3+1)^2}}=\color{blue}{\bf 1876}\;\rm nm$$
$$\lambda_{\rm min}=\dfrac{91.18}{\dfrac{1}{3^2}-\dfrac{1}{(\infty)^2}}=\color{blue}{\bf 820}\;\rm nm$$
We can stop here since the maximum and the minimum values are greater than the maximum wavelength of the visible light where we know that the visible light ranges from 400 nm to 700 nm.
But let's take an extra one to make sure.
$\bullet$ For $m=4$, using (1),
$$\lambda_{\rm max}=\dfrac{91.18}{\dfrac{1}{4^2}-\dfrac{1}{(4+1)^2}}=\color{blue}{\bf 4052}\;\rm nm$$
$$\lambda_{\rm min}=\dfrac{91.18}{\dfrac{1}{4^2}-\dfrac{1}{(\infty)^2}}=\color{blue}{\bf 1459}\;\rm nm$$
From the 4 $m$s there, we can see that the visible light occurs at $m=2$ starting from $n=2+1=3$ and so on...
Now we can calculate the visible light wavelengths that occur at $m=2$ as follows;
$$\lambda_{3\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{(2+1)^2}}=\color{red}{\bf 656.5}\;\rm nm$$
$$\lambda_{4\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{4^2}}=\color{red}{\bf 486.3}\;\rm nm$$
$$\lambda_{5\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{5^2}}=\color{red}{\bf 434.2}\;\rm nm$$
$$\lambda_{6\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{6^2}}=\color{red}{\bf 410.3}\;\rm nm$$
$$\lambda_{7\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{7^2}}=\color{blue}{\bf 397.1}\;\rm nm$$
We can stop here since this is out of the range of visible light.
But to make sure, let's take an extra one..
$$\lambda_{8\rightarrow2}=\dfrac{91.18}{\dfrac{1}{2^2}-\dfrac{1}{8^2}}=\color{blue}{\bf 389}\;\rm nm$$
Therefore, we have only four visible wavelengths in the hydrogen spectrum which are:
$$\boxed{\rm \color{red}{\bf 656.5}\;nm,\;\color{red}{\bf 486.3}\;nm,\; \color{red}{\bf 434.2}\;nm,\; \color{red}{\bf 410.3}\;nm }$$
