If E_H represents the total energy of an electron in the second...

The total energy of an electron in the $n$-th Bohr orbit of a hydrogen-like atom is given by $E = -13.6 \dfrac{Z^2}{n^2} \text{ eV}$.

For the hydrogen atom ($Z = 1$) in the second Bohr orbit ($n = 2$):
$E_H = -13.6 \times \dfrac{1^2}{2^2} = -\dfrac{13.6}{4} \text{ eV}$

For the singly ionized helium atom ($He^+$, $Z = 2$) in the third Bohr orbit ($n = 3$):
$E_{He} = -13.6 \times \dfrac{2^2}{3^2} = -13.6 \times \dfrac{4}{9} \text{ eV}$

Taking the ratio of the two energies:
$\dfrac{E_H}{E_{He}} = \dfrac{-\dfrac{13.6}{4}}{-13.6 \times \dfrac{4}{9}} = \dfrac{1}{4} \times \dfrac{9}{4} = \dfrac{9}{16}$

The ratio $E_H : E_{He}$ is $9 : 16$.

Answer: $9 : 16$