Let the length of shorter tube be \(L_1\) and that of the longer tube be \(L_2\).
Since water in filled in it, tube behaves as a closed organ pipe in both arms.
Modes of vibration of air column in a closed organ pipe are given by
\(f_n = \dfrac{(2n-1)v}{4L}~~~~~\text{where}~n=1,2,3,4,5......\)
For shorter arm:
From the data given in the question,
\(550 = \dfrac{330}{4\times L_1}\Rightarrow L_1 = \dfrac{33}{4\times 55} = 0.15~\text{m}~\text{or}~15~\text{cm}\)
Similarly, for longer arm:
From the data given in the question,
\(550=\dfrac{3\times 330}{4\times L_2} \Rightarrow L_2 =\dfrac{3\times 330}{4\times 550}=0.45~\text{m}~\text{or}~45~\text{cm}\)
Thus, option (b) is the correct answer.