Let the length of the closed organ pipe be $L_c = 20 \text{ cm}$. The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4L_c}$. Let the length of the open organ pipe be $L_o$. The second overtone of an open organ pipe corresponds to the third harmonic. Its frequency is given by $f_o = \frac{3v}{2L_o}$. According to the question, $f_c = f_o$: $\frac{v}{4L_c} = \frac{3v}{2L_o}$ Rearranging to solve for $L_o$: $L_o = \frac{3v \times 4L_c}{2v} = 6L_c$ Given $L_c = 20 \text{ cm}$: $L_o = 6 \times 20 \text{ cm} = 120 \text{ cm}$