Step 1: Work with photon energy and atomic energy levels, instead of the Rydberg wavelength formula directly.
The electron's energy in level $n$ of hydrogen is $E_n = -\frac{13.6}{n^2}$ eV. A spectral line's photon energy is the gap between the starting and ending levels.
Step 2: Identify which transition gives the shortest wavelength in the Balmer series.
The Balmer series always lands on $n=2$. Shorter wavelength means higher photon energy, and photon energy is largest for the transition starting at the highest possible level, $n \to \infty$.
Step 3: Compute this energy gap.
\[ \Delta E = E_{\infty} - E_2 = 0 - \left(-\frac{13.6}{4}\right) = 3.4\ \text{eV} \]
Step 4: Convert to a wavelength using $\lambda = \frac{hc}{\Delta E}$, with $hc = 1240$ eV nm.
\[ \lambda = \frac{1240}{3.4} \approx 364.7\ \text{nm} \]
Step 5: Match with the closest option.
This rounds to $364.6$ nm, the small gap coming from rounding $hc$ to $1240$ eV nm and the ground state energy to $13.6$ eV.
Final Answer:
\[ \boxed{364.6\ \text{nm}} \]