Frequency of $1$ st harmonic of $AB$

$=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}$

Frequency of $2$ nd harmonic of $\mathrm{CD}$

$=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}$

Given that the two frequencies are equal.

$\therefore \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{AB}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{CD}}}{\mathrm{m}}}$

$\Rightarrow \frac{\mathrm{T}_{\mathrm{AB}}}{4}=\mathrm{T}_{\mathrm{CD}} \Rightarrow \mathrm{T}_{\mathrm{AB}}=4 \mathrm{T}_{\mathrm{CD}}$             $…(i)$

For rotational equilibrium of massless rod, taking torque about point $O.$

$\mathrm{T}_{\mathrm{AB}} \times \mathrm{x}=\mathrm{T}_{\mathrm{CD}}(\mathrm{L}-\mathrm{x})$             $…(ii)$

For translational equilibrium,

$\mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=\mathrm{mg}$       $..(iii)$

On solving, $(i)$ $\&(iii)$ we get, $\mathrm{T}_{\mathrm{CD}}-\frac{\mathrm{mg}}{5}$

$\therefore \mathrm{T}_{\mathrm{AB}}=\frac{4 \mathrm{mg}}{5}$

Substituting these values in $(ii)$ we get

$\frac{4 m g}{5} \times x=\frac{m g}{5}(L-x)$

$\Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{L}}{5}$