As given that $\mathrm{C}_1$ is circular coil of radius $\mathrm{R}$, length $\mathrm{L}$, number of turns per unit length
$$
\mathrm{n}_1=\frac{\mathrm{L}}{2 \pi \mathrm{R}}
$$
$\mathrm{C}_2$ is square of side a and perimeter $\mathrm{L}$, number of turns per unit length $\mathrm{n}_2=\frac{\mathrm{L}}{4 \mathrm{a}}$



So magnetic moment of $\mathrm{C}_1$
$$
\mathrm{m}_1=\mathrm{n}_1 \mathrm{IA}_1=\frac{\text { L.I. } \pi \mathrm{R}^2}{2 \pi \mathrm{R}}
$$
And magnetic moment of $\mathrm{C}_2$
$$
\begin{aligned}
\mathrm{m}_2=\mathrm{n}_2 \mathrm{IA}_2 &=\frac{\mathrm{L}}{4 \mathrm{a}} \cdot \mathrm{I}^2 \mathrm{a}^2 \\
\mathrm{~m}_1 &=\frac{\mathrm{LIR}}{2} \\
\mathrm{~m}_2 &=\frac{\text { LIa }}{4}
\end{aligned}
$$
Moment of inertia of $\mathrm{C}_1$
$$
\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}
$$
Moment of inertia of $\mathrm{C}_2$
$$
\Rightarrow \quad \quad \mathrm{I}_2=\frac{\mathrm{Ma}^2}{12}
$$
As the frequencies for both coil are given by
$$
\begin{aligned}
f_1 &=f_2 \\
\left(\frac{2 \pi}{T_1}\right) &=\left(\frac{2 \pi}{T_2}\right),\left(T_1=T_2\right)
\end{aligned}
$$
So, time period of $\mathrm{C}_1$
$$
\Rightarrow \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{I}_1}{\mathrm{~m}_1 \mathrm{~B}}}
$$
And time period of $\mathrm{C}_2$
$$
\begin{aligned}
&\Rightarrow \quad \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{I}_2}{\mathrm{~m}_2 \mathrm{~B}}} \\
&\text { From }(v): \quad T_1=T_2 \\
&2 \pi \sqrt{\frac{\mathrm{I}_1}{\mathrm{~m}_1 \mathrm{~B}}}=2 \pi \sqrt{\frac{\mathrm{I}_2}{\mathrm{~m}_2 \mathrm{~B}}} \\
&\frac{\mathrm{I}_1}{\mathrm{~m}_1}=\frac{\mathrm{I}_2}{\mathrm{~m}_2} \\
&\frac{\mathrm{m}_2}{\mathrm{~m}_1}=\frac{\mathrm{I}_2}{\mathrm{I}_1} \\
&
\end{aligned}
$$
Putting the values by Eqs. (i), (ii), (iii) and (iv) in (vi)
So,
$$
\begin{aligned}
\frac{\mathrm{LIa} \cdot 2}{4 \times \mathrm{LIR}} &=\frac{\mathrm{Ma}^2 \cdot 2}{12 \cdot \mathrm{MR}^2} \\
\frac{\mathrm{a}}{2 \mathrm{R}} &=\frac{\mathrm{a}^2}{6 \mathrm{R}^2} \\
\mathrm{a} &=3 \mathrm{R}
\end{aligned}
$$