What is the ratio of the ionization energy of a hydrogen atom from...

The ionization energy from the $n^{\text{th}}$ state of a hydrogen-like species is given by $IE = 13.6 \times \dfrac{Z^2}{n^2} \mathrm{eV}$.

For a hydrogen atom ($Z=1$) in its first excited state ($n=2$):
$IE_1 = 13.6 \times \dfrac{1^2}{2^2} = \dfrac{13.6}{4} \mathrm{eV}$.

For a $\mathrm{He}^{+}$ ion ($Z=2$) in its second excited state ($n=3$):
$IE_2 = 13.6 \times \dfrac{2^2}{3^2} = 13.6 \times \dfrac{4}{9} \mathrm{eV}$.

The ratio of their ionization energies is:
$\dfrac{IE_1}{IE_2} = \dfrac{1/4}{4/9} = \dfrac{9}{16}$.

Thus, the required ratio is $9:16$.

Answer: $9:16$