Answer
$3.4~eV~~$ of energy is required to ionize a hydrogen atom that is in its first excited state.
Work Step by Step
When a hydrogen atom is in its first excited state, then $n = 2$
We can find the energy when $n = 2$:
$E_n = -\frac{13.6~eV}{n^2}$
$E_2 = -\frac{13.6~eV}{2^2}$
$E_2 = -3.4~eV$
$3.4~eV~~$ of energy is required to ionize a hydrogen atom that is in its first excited state.
