Doubt in a question related to electronic transitions in Bohr's orbits
Since only one answer is correct, we must consider the transitions in the visible region only i.e., the Balmer Series. If multiple answers are correct then it is wise to choose both the options B and C.

The transition $n_2 \to n_1$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.

$$n_2=n_1 + n$$
where $n$ represents the $n^{th}$ line in a particular series. Since here we are talking about Balmer series, $n_1=2$ and for the third line $n=3$. Substituting the values
$$n_2=2+3=5$$

Hence the required transition is given by option B.