In hydrogen atom the kinetic energy of electron is 3.4 eV. The distance of that electron from the nucleus

In H-atom the kinetic energy of electron is 3.4 ev

Then total energy of electron E_total = −K.E
So, E_total = −(3.4eV)

so the electron in first exited state n=2

The distance of that electron from the nucleus (r) =0.529 $\times$ n²/Z

(r) =0.529 $\times$2²/1 

(r) = 2.116 A⁰