The wavelength ($\lambda$) of a spectral line is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.

1. Hydrogen atom (Z=1): The first line of the Balmer series corresponds to the transition from $n_2=3$ to $n_1=2$. $\frac{1}{\lambda_H} = R(1)^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$. Given $\lambda_H = 6561 \mathring{A}$.

2. Singly ionized Helium atom (He$^+$, Z=2): The second line of the Balmer series corresponds to the transition from $n_2=4$ to $n_1=2$. $\frac{1}{\lambda_{He}} = R(2)^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = 4R\left(\frac{3}{16}\right) = \frac{3R}{4}$.

3. Ratio: $\frac{\lambda_{He}}{\lambda_H} = \frac{5R/36}{3R/4} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27}$. $\lambda_{He} = 6561 \times \frac{5}{27} = 243 \times 5 = 1215 \mathring{A}$.