To solve this problem, we need to understand the relationship between the frequencies of closed and open organ pipes.$\\$For a closed organ pipe, the fundamental frequency is given by:$\\ f_{closed} = \frac{v}{4L_{closed}} \\$where $L_{closed} = 20$ cm$= 0.2$ m and $v$ is the speed of sound in air.$\\$For an open organ pipe, the frequency of the second overtone (third harmonic) is given by:$\\ f_{open, 2nd\ overtone} = \frac{3v}{2L_{open}} \\$According to the problem, the fundamental frequency of the closed pipe is equal to the second overtone of the open pipe:$\\ \frac{v}{4 \times 0.2} = \frac{3v}{2L_{open}} \\$Simplify and solve for $L_{open}: \\ \frac{v}{0.8} = \frac{3v}{2L_{open}} \\$Cancel $v$ from both sides:$\\ \frac{1}{0.8} = \frac{3}{2L_{open}} \\$Cross-multiply to solve for $L_{open}: \\ 2L_{open} = 3 \times 0.8 \\ 2L_{open} = 2.4 \\ L_{open} = \frac{2.4}{2} \\ L_{open} = 1.2$ m$\\ L_{open} = 120$ cm$\\$Therefore, the length of the organ pipe open at both ends is $120$ cm.$\\$This corresponds to Option 2.