For a closed organ pipe, the fundamental frequency is $f_{c} = \frac{v}{4L_c}$. The overtones are odd harmonics. The $n^{\text{th}}$ overtone corresponds to the $(2n+1)^{\text{th}}$ harmonic. Frequency of the $4^{\text{th}}$ overtone of the closed pipe = $(2 \times 4 + 1) f_c = 9 \left(\frac{v}{4L_c}\right)$.

For an open organ pipe, the fundamental frequency is $f_{o} = \frac{v}{2L_o}$. The overtones are integer harmonics. The $n^{\text{th}}$ overtone corresponds to the $(n+1)^{\text{th}}$ harmonic. Frequency of the $3^{\text{rd}}$ overtone of the open pipe = $(3 + 1) f_o = 4 \left(\frac{v}{2L_o}\right) = \frac{2v}{L_o}$.

Given that these two frequencies are equal: $\frac{9v}{4L_c} = \frac{2v}{L_o}$ $\frac{L_c}{L_o} = \frac{9}{8}$.

Therefore, the ratio of the length of the closed pipe to the length of the open pipe is $9:8$.