The angular momentum of an electron in hydrogen atom $=\frac{h}{\pi}$
As we know, from Bohr’s second postulate,
$mvr=\frac{h}{\pi} \Rightarrow v=\frac{h}{\pi mr}$...(i)
Now, kinetic energy of electron
$=\frac{1}{2}mv^{2}=\frac{1}{2}m\times\left(\frac{h}{\pi mr}\right)^{2}$
[from Eq. $\left(i\right)$]
$=\frac{h^{2}}{2\pi^{2} mr^{2}}$
$=\frac{\left(6.63\times10^{-34}\right)^{2}}{2\times\left(3.14\right)^{2}\times9.1\times10^{-31} \times\left(5.3\times10^{-11}\right)^{2}\times1.6\times10^{-19}}$
$=3.4 \,eV$