Bohr Model: Energy of the Electron

Total energy in the $n$th orbit: $E_n = \dfrac{-13.6}{n^2}$ eV
The first excited state corresponds to $n = 2$.
Energy in the first excited state $= \dfrac{-13.6}{2^2} = -3.4$ eV
In a Bohr orbit, K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
$\therefore$ K.E. $= -(-3.4) = +3.4$ eV

In a Bohr orbit the kinetic energy is $K = \dfrac{1}{2}\dfrac{ke^2}{r}$ and the potential energy is $U = -\dfrac{ke^2}{r}$.
So the total energy is $E = K + U = -\dfrac{1}{2}\dfrac{ke^2}{r} = -K$.
Thus K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
K.E. $= -(-3.4) = +3.4$ eV
(The potential energy in this state is $-6.8$ eV.)