Explanation:
Frequency of the second overtone (i.e. third harmonic) of the open pipe is
\(f = \frac{{3v}}{{2{L_0}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
where \(v\) is the speed of sound in air and \({L_0}\) is the length of the open pipe.
Frequency of the first overtone (i.e. third harmonic) of the closed pipe is
\({f^\prime } = \frac{{3v}}{{4{L_C}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As per question, \(f = {f^\prime }\)
\(\therefore \frac{{3v}}{{2{L_0}}} = \frac{{3v}}{{4{L_C}}} \Rightarrow \frac{{{L_0}}}{{{L_C}}} = \frac{2}{1}\)