The total energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -13.6/n^2\,\text{eV}$. The energy of the electron in the first excited state ($n = 2$) is:

The first excited state corresponds to $n = 2$, so $E_2 = -13.6 / 2^2 = -13.6/4 = -3.4\,\text{eV}$. The ground state $n = 1$ has energy $-13.6\,\text{eV}$; the second excited state $n = 3$ has $-1.51\,\text{eV}$.