Consider an electron in the second stationary orbit of a He^ + ion...

The radius of the $n^{\text{th}}$ Bohr orbit for a hydrogen-like species is given by:
$r \propto \dfrac{n^2}{Z}$

For the $\text{He}^{+}$ ion in the second orbit ($n=2, Z=2$):
$r_{\text{He}^{+}} \propto \dfrac{2^2}{2} = 2$

For the hydrogen atom in the first orbit ($n=1, Z=1$):
$r_{\text{H}} \propto \dfrac{1^2}{1} = 1$

The ratio of their radii is:
$\dfrac{r_{\text{He}^{+}}}{r_{\text{H}}} = \dfrac{2}{1}$

Assuming the orbits to be circular, the area enclosed by an orbit is $A = \pi r^2$.
Therefore, the ratio of the areas is proportional to the square of the ratio of their radii:
$\dfrac{A_{\text{He}^{+}}}{A_{\text{H}}} = \left( \dfrac{r_{\text{He}^{+}}}{r_{\text{H}}} \right)^2 = \left( \dfrac{2}{1} \right)^2 = \dfrac{4}{1}$

The required ratio is $4:1$.

Answer: $4:1$