Answer
${\bf 0.5}\;\rm m$
Work Step by Step
Recalling that the inductance is given by
$$L=\dfrac{\mu_0N^2A}{l}$$
$$L=\dfrac{\mu_0N^2\pi r_{\rm sol}^2}{l}\tag 1$$
The inductor length is given by
$$l=ND_{\rm wire}\tag 2$$
where $D_{\rm wire}$ is the diameter of the wire.
So,
$$N=\dfrac{ l}{ D_{\rm wire}}$$
Plug into (1),
$$L=\dfrac{ \mu_0l^2\pi r_{\rm sol}^2}{ D^2_{\rm wire}l} $$
$$L=\dfrac{ \mu_0l \pi r_{\rm sol}^2}{ D^2_{\rm wire} } $$
Solving for $l$;
$$l=\dfrac{ D^2_{\rm wire}L } { \pi \mu_0 r_{\rm sol}^2}\tag 3$$
Recalling that
$$\omega=2\pi f=\dfrac{1}{\sqrt{LC}}$$
Hence,
$$L=\dfrac{1}{4\pi^2f^2C}$$
Plug into (3),
$$l=\dfrac{ D^2_{\rm wire} } { \mu_0 \pi r_{\rm sol}^2(4\pi^2f^2C) } $$
$$l=\dfrac{ D^2_{\rm wire} } {4\mu_0 \pi^3 r_{\rm sol}^2 f^2C } $$
Plug the known;
$$l=\dfrac{ (0.25\times 10^{-3})^2 } {4(4\pi\times 10^{-7})\pi^3 (0.02)^2 (1)^2(1) } =\bf 1\;\rm m$$
And since the wire is wrapped on the cylinder in 2 layers, so
$$l_{\rm sol}=\dfrac{l}{2}=\dfrac{1}{2}$$
$$l_{\rm sol}=\color{red}{\bf 0.5}\;\rm m$$
