Step 1: For the Balmer series, $n_1 = 2$. The longest wavelength transition corresponds to the minimum energy transition, which occurs at the lowest possible value of $n_2$. Thus, $n_2 = 3$.

Step 2: Using the Rydberg formula for wavenumber: $\bar{\nu} = 109,677 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ cm}^{-1}$.

Step 3: Substitute $n_1 = 2$ and $n_2 = 3$ into the formula: $\bar{\nu} = 109,677 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \text{ cm}^{-1}$.

Step 4: Calculate the values inside the bracket: $\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$.

Step 5: Multiply by the Rydberg constant: $\bar{\nu} = 109,677 \times \frac{5}{36} \text{ cm}^{-1} = 15232.92 \text{ cm}^{-1}$, which can be written as $1.523 \times 10^4 \text{ cm}^{-1}$.