$\nu_{A}=\frac{2 v}{2 l_{A}}=\frac{v}{l_{A}}$$.........(1)$

The first overtone of a close organ pipe $\mathrm{B}$ is

$\nu_{B}=\frac{3 v}{4 l_{B}}$$............(2)$

Dividing $(1)$ by $(2)$ we get

$\frac{\nu_{A}}{\nu_{B}}=\frac{v}{l_{A}} \times \frac{4 l_{B}}{3 v}$

$\frac{\nu_{A}}{\nu_{B}}=\frac{4 l_{B}}{3 l_{A}}$  $.............(3)$

The second overtone of an open organ pipe $A$ is

$\nu_{A}=\frac{3 v}{2 l_{A}}$

The second overtone of a close organ pipe $B$ is

$\nu_{B}=\frac{5 v}{4 l_{B}}$

since, the second overtone of an open organ pipe $\mathrm{A}$ and a closed pipe $\mathrm{B}$ have the same frequency at a given temperature,

$\nu_{A}=\nu_{B}$

$\frac{3 v}{2 l_{A}}=\frac{5 v}{4 l_{B}}$

$\frac{l_{B}}{l_{A}}=\frac{10}{12}=\frac{5}{6}$$..........(4)$

Now, using equation $(4)$ in above equation $(3)$ we get

$\frac{\nu_{A}}{\nu_{B}}=\frac{4 \times 5}{3 \times 6}$

$\frac{\nu_{A}}{\nu_{B}}=\frac{20}{18}=\frac{10}{9}$

Hence, lengths of $B$ to that of $A$ is $5: 6$ and frequencies of first overtone of $A$$\&$ B is $10: 9$