When an e transition from n$^{th}$ to ground state then $$\sum {(n - 1)} $$ spectral lines are produced.

Given, $n=4$

So, $$\sum {(4 - 1) = \sum {3 = 3 + 2 + 1 = 6} } $$

Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the $4^{th}$ energy level