Solution

We write Newton's second law for the weight, considering the elastic force of the spring $F_{\text{el}}$:
$$m a_{\text{cp}} = F_{\text{el}} = k\Delta x$$

Let's express $\Delta x$ through the length of the undeformed spring $l$:
$$m a_{\text{cp}} = F_{\text{el}} = k(R - l) \quad (1)$$

Let's find the centripetal acceleration through the angular velocity of rotation $\omega$:
$$a_{\text{cp}} = \omega^2 R$$

Substitute this into (1):
$$m \omega^2 R = k(R - l)$$
$$\frac{k}{m}(R - l) = \omega^2 R$$

From which we find $l$:
$$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$