When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

The wavelength of line in case of Balmer series is given by
$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$ , where n = 3, 4, 5…
and R = Rydberg constant
So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series.