Answer
$4$
Work Step by Step
We know,
$E_2-E_1=-13.6\Big (\frac{1}{n_2^2}-\frac{1}{n_2^2}\Big )\;eV$
In the the previous section, we have obtained
$E_2-E_1=2.55\;eV\\
or, -13.6\Big (\frac{1}{n_2^2}-\frac{1}{n_2^2}\Big )=2.55\;eV\\
or, \Big (\frac{1}{n_2^2}-\frac{1}{n_2^2}\Big )=-\frac{2.55}{13.6}\\
or,\Big (\frac{1}{n_2^2}-\frac{1}{n_2^2}\Big )=-\frac{3}{16}\\
or,\Big (\frac{1}{n_2^2}-\frac{1}{n_2^2}\Big )=\Big (\frac{1}{4^2}-\frac{1}{2^2}\Big )$
Comparing we obtain
$n_2=4$ and $n_1=2$
Therefore, the higher quantum number in this transition is $4$
