Answer
${\bf 4.16}\;\rm eV$
Work Step by Step
We know that the minimum energy required to ionize a hydrogen atom is
$$\Delta E_{\rm atom}=13.6 \;\rm eV$$
Now we need to find the energy of the light photon
$$ E_{\rm photon}=\dfrac{hc}{e\lambda}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})(70\times 10^{-9})}=\bf 17.76\;\rm eV$$
Now, we can determine the kinetic energy of the freed electron by subtracting the ionization energy of the atom from the total energy of the photon.
$$K=E_{\rm photon}-\Delta E_{\rm atom}=17.76-13.6$$
$$K=\color{red}{\bf 4.16}\;\rm eV$$
