Let the length of the open organ pipe be $L_o$ and the length of the closed organ pipe be $L_c$. The fundamental frequency of an open organ pipe is given by $f_o = \frac{v}{2L_o}$. The frequency of the third harmonic (first overtone) of a closed organ pipe is given by $f_c = \frac{3v}{4L_c}$. According to the question, $f_o = f_c$: $\frac{v}{2L_o} = \frac{3v}{4L_c}$ Rearranging to solve for $L_o$: $L_o = \frac{4L_c}{6} = \frac{2L_c}{3}$ Given $L_c = 20 \text{ cm}$, $L_o = \frac{2 \times 20}{3} = \frac{40}{3} = 13.33 \text{ cm}$ The calculated length is $13.33 \text{ cm}$, and the closest available option provided in the question is $13.2 \text{ cm}$.