Answer
$\approx 43\;\rm loops$
Work Step by Step
First of all, we need to find $L$, so we can find the number of loops $N$.
We know that the frequency of a resonant circuit is given by
$$f_0=\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}}$$
Solving for $L$;
$$(2\pi f_0 )^2= \dfrac{1}{LC} $$
$$L = \dfrac{1}{(2\pi f_0 )^2 C}\tag 1 $$
We also know that $$L=\dfrac{\mu_0N^2A}{l}$$
Plugging into (1);
$$\dfrac{\mu_0N^2A}{l_{solenoid}} = \dfrac{1}{(2\pi f_0 )^2 C} $$
Noting that the length of the air-core solenoid conductor is given by the number of loops times the diameter of the wire. so $l_{solenoid}= ND_{wire}$
Thus,
$$\dfrac{\mu_0N^2A}{ND_{wire}} = \dfrac{1}{ 4\pi^2 f_0^2 C} $$
The cross-sectional area of the solenoid is a circle, so its area is given by $A=\pi r^2$ where $r$ is the radius of one loop which is the radius of the solenoid itself.
$$\dfrac{\mu_0N \pi r_{loop}^2 }{ D_{wire}} = \dfrac{1}{4\pi^2 f_0^2 C} $$
Solving for $N$;
$$ N = \dfrac{D_{wire}}{4\mu_0\pi^3 f_0^2 r_{loop}^2 C} \tag 2$$
Recall that the number of loops is also given by the length of the wire divided by the circumference of one loop; $N=l_{wire}/2\pi r_{loop}$; thus
$$r_{loop}=\dfrac{l_{wire}}{2\pi N}$$
Plugging into (2);
$$ N = \dfrac{D_{wire}}{4\mu_0\pi^3 f_0^2\left(\dfrac{l_{wire}}{2\pi N}\right)^2 C} = \dfrac{4\pi ^2 N^2D_{wire}}{4\mu_0\pi^3 f_0^2l_{wire}^2 C} $$
$$ 1 = \dfrac{ N D_{wire}}{ \mu_0\pi f_0^2l_{wire}^2 C} $$
Therefore,
$$N= \dfrac{ \mu_0\pi f_0^2l_{wire}^2 C}{ D_{wire}} $$
Plugging the known;
$$N=\dfrac{ 4\pi^2\cdot 10^{-7}\cdot (18000)^2\cdot (12)^2\cdot 260\times10^{-9} }{ 1.1\times10^{-3}}=43.5\;\rm loop$$
$$N\approx \color{red}{\bf 43}\;\rm loop$$
