Solution

Since the ball does not float, the vertical Archimedean force is completely compensated by the vertical component of the thread tension force $$T\cos\alpha =\frac{4\pi r^3}{3}\rho g$$ From where we could find the thread tension force $$T=\frac{4\pi r^3}{3\cos\alpha}\rho g\tag{1}$$ Alternatively, we could write an equilibrium condition for the Horizontal Archimedean force , which is compensated by the horizontal force of the thread tension $$T\sin\alpha = \rho V\omega^2x\tag{2}$$ It can be seen from the figure that the ball's center of mass is at a distance $l+r$ from the point of attachment of the thread: $$x=R-(l+r)\sin\alpha$$ After substituting into $(2)$ we obtain $$\rho\frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = T\sin\alpha\tag{3}$$ Substituting $(1)$ into $(3)$

$$\rho\frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = \frac{4\pi r^3}{3\cos\alpha}\rho g\sin\alpha$$

From where we find the angular velocity $$\boxed{\omega = \sqrt{\frac{g\tan\alpha}{R-(l+r)\sin\alpha}}}$$