One end of a massless spring of spring constant k and natural...

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude $\dfrac{{v}^{2}}{r}$. The horizontal force on the particle is due to the spring and is given by $kx$, where $x$ is the elongation and $k$ is the spring constant.

$kx = \dfrac{m{v}^{2}}{r} = m{\omega}^{2}r = m{\omega}^{2}\left({l}_{0} + x\right)$

$⇒ \left(k - m{\omega}^{2}\right)x = m{\omega}^{2}{l}_{0}$

$⇒ x = \dfrac{m{\omega}^{2}{l}_{0}}{k - m{\omega}^{2}}$