Consider the following spectral lines for atomic hydrogen : A....

The energy of a spectral line in the hydrogen atom is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ eV.
A. First line of Paschen series: $n_1 = 3, n_2 = 4$.
$E_A = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \times \frac{7}{144} \approx 0.66$ eV.

B. Second line of Balmer series: $n_1 = 2, n_2 = 4$.
$E_B = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \times \frac{3}{16} = 2.55$ eV.

C. Third line of Paschen series: $n_1 = 3, n_2 = 6$.
$E_C = 13.6 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{36} \right) = 13.6 \times \frac{3}{36} = 13.6 \times \frac{1}{12} \approx 1.13$ eV.

D. Fourth line of Brackett series: $n_1 = 4, n_2 = 8$.
$E_D = 13.6 \left( \frac{1}{4^2} - \frac{1}{8^2} \right) = 13.6 \left( \frac{1}{16} - \frac{1}{64} \right) = 13.6 \times \frac{3}{64} \approx 0.6375$ eV.

Comparing the values: $E_D (0.6375) < E_A (0.66) < E_C (1.13) < E_B (2.55)$.
The ascending order of energy is $D < A < C < B$.