Consider a circuit consisting of a capacitor of capacitance C and a...

Let the larger coil be coil 1 and the smaller coil be coil 2.

The self-inductance of the larger coil is given by:
$L_1 = \mu_0 n_1^2 A_1 l_1 = \mu_0 N^2 S d = L$

The self-inductance of the smaller coil is:
$L_2 = \mu_0 n_2^2 A_2 l_2 = \mu_0 (2N)^2 \left(\dfrac{S}{2}\right) \left(\dfrac{d}{2}\right) = \mu_0 (4N^2) \left(\dfrac{Sd}{4}\right) = \mu_0 N^2 S d = L$

The mutual inductance between the two coils is:
$M = \mu_0 n_1 n_2 A_{\text{common}} l_{\text{common}}$
Since the smaller coil is completely inside the larger coil, the common area is $S/2$ and the common length is $d/2$.
$M = \mu_0 (N) (2N) \left(\dfrac{S}{2}\right) \left(\dfrac{d}{2}\right) = \dfrac{1}{2} \mu_0 N^2 S d = \dfrac{L}{2}$

Let $i_1$ be the current in the larger coil and $i_2$ be the current in the smaller coil. The voltage across the larger coil is:
$V = L_1 \dfrac{di_1}{dt} + M \dfrac{di_2}{dt}$

Since the smaller coil is short-circuited, the net voltage across it is zero:
$0 = L_2 \dfrac{di_2}{dt} + M \dfrac{di_1}{dt} \Rightarrow \dfrac{di_2}{dt} = -\dfrac{M}{L_2} \dfrac{di_1}{dt}$

Substituting this into the first equation, we get the effective inductance $L_{\text{eq}}$:
$V = L_1 \dfrac{di_1}{dt} - \dfrac{M^2}{L_2} \dfrac{di_1}{dt} = \left( L_1 - \dfrac{M^2}{L_2} \right) \dfrac{di_1}{dt}$
$L_{\text{eq}} = L_1 - \dfrac{M^2}{L_2} = L - \dfrac{(L/2)^2}{L} = L - \dfrac{L}{4} = \dfrac{3L}{4}$

The resonant angular frequency of the circuit is:
$\omega = \dfrac{1}{\sqrt{L_{\text{eq}} C}} = \dfrac{1}{\sqrt{\dfrac{3L}{4} C}} = \dfrac{2}{\sqrt{3LC}}$

Answer: $\dfrac{2}{\sqrt{3\, LC}}$