As the hydrogen atoms emit radiation of six different wavelengths, some of them must have been excited to $\text{n}=4$, The energy in $\text{n}=4$ state is

 ${\text{E}}_4=\frac{{\text{E}}_1}{4^2}=-\frac{13·6\text{e}\text{V}}{16}=-0·85\text{eV.}$

The energy needed to take a hydrogen atom from its ground state to $\text{n}=4$ is

$\text{13}\text{.6 eV - 0}\text{.85 eV = 12}\text{.75 eV}$ .

The photons of the incident radiation should have $12.75\text{eV}$ of energy. So

 $\frac{\text{hc}}{\lambda }=12·75\text{eV}$

or , $\lambda =\frac{\text{hc}}{12·75\text{eV}}$

 $=\frac{1242\text{eV}\text{nm}}{12·75\text{eV}}$

 $\text{= 97}\text{.5 nm}$.