For ground level, $n_{1}=1$

Let $E_{1}$ be the energy of this level. It is known that $E_{1}$ is related with $n_{1}$ as

$E_{1}=\frac{-13.6}{n_{1}^{2}}\, e V$

$=\frac{-13.6}{1^{2}}=-13.6\, e V$

The atom is excited to a higher level, $n_{2}=4$

Let $E_{2}$ be the energy of this level.

$\therefore E_{2}=\frac{-13.6}{n_{2}^{2}} \,e V$

$=\frac{-13.6}{4^{2}}=-\frac{13.6}{16}\, e V$

The amount of energy absorbed by the photon is given as

$E=E_{2}-E_{1}$

$=\frac{-13.6}{16}-\left(-\frac{13.6}{1}\right)$

$=\frac{13.6 \times 15}{16} \,e V$

$=\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} \,J$

For a photon of wavelength $\lambda,$ the expression of energy is written as

$E=\frac{h c}{\lambda}$

Where, $h=$ Planck's constant $=6.6 \times 10^{-34} \,Js$

$c=$ Speed of light $=3 \times 10^{8} \,m / s$

$\therefore \lambda=\frac{h c}{E}$

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}}$

$=9.7 \times 10^{-8}\, m =97\, nm$

And, frequency of a photon is given by the relation, $v=\frac{c}{\lambda}$

$=\frac{3 \times 10^{8}}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \,Hz$

Hence, the wavelength of the photon is $97\, nm$ while the frequency is $3.1 \times 10^{15} \,Hz$.