For closed organ pipe: Frequency of different modes, $\nu_{n}^{\prime}=\frac{(2 p-1) v}{4 L^{\prime}}$ where $(2 p-1)$ represents nth harmonic.

Thus frequency of nth harmonic in closed pipe, $\nu_{n}^{\prime}=\frac{n v}{4 L^{\prime}}$

Ist overtone i.e $p=2, \nu_{3}^{\prime}=\frac{3 v}{4 L^{\prime}}$

For open organ pipe: Frequency of mth harmonic $\nu_{m}=\frac{m v}{2 L}$

Ist overtone i.e $m=2, \nu_{2}=\frac{2 v}{2 L}$

Given $: \nu_{3}^{\prime}=\nu_{2}$

$\frac{3 v}{4 L^{\prime}}=\frac{v}{L} \Longrightarrow \frac{L^{\prime}}{L}=\frac{3}{4}…..(1)$

Also given, $\nu_{n}^{\prime}=\nu_{m}$

$\frac{n v}{4 L^{\prime}}=\frac{m v}{2 L}$

$\frac{n}{2}=m \frac{L^{\prime}}{L} \Longrightarrow \frac{n}{2}=m \frac{3}{4}$

Thus $2 n=3 m$

Now $n=9$ and $m=6$ satisfies the relation.