1. Let the mass and radius of the solid sphere be $m$ and $R$.

2. Since the sphere is rolling without slipping, $v = \omega R$ where $v$ is the speed of the centre of mass and $\omega$ is the angular speed.

3. Angular momentum about the axis of rotation: For a solid sphere, moment of inertia about its own axis is $I = \frac{2}{5}mR^2$ So angular momentum about the axis of rotation is $L = I\omega = \frac{2}{5}mR^2\omega$

4. Total energy of the moving sphere: This is the sum of translational and rotational kinetic energies. $E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ Using $v=\omega R$ and $I=\frac{2}{5}mR^2$, $E = \frac{1}{2}m(\omega R)^2 + \frac{1}{2}\cdot \frac{2}{5}mR^2\omega^2$ $E = \frac{1}{2}mR^2\omega^2 + \frac{1}{5}mR^2\omega^2$ $E = \frac{7}{10}mR^2\omega^2$

5. Given ratio: $L : E = \pi : 22$ So, $\frac{L}{E} = \frac{\pi}{22}$ Substitute $L$ and $E$: $\frac{\frac{2}{5}mR^2\omega}{\frac{7}{10}mR^2\omega^2} = \frac{\pi}{22}$ Cancel $mR^2$: $\frac{2}{5}\cdot \frac{10}{7}\cdot \frac{1}{\omega} = \frac{\pi}{22}$ $\frac{4}{7\omega} = \frac{\pi}{22}$

6. Solve for $\omega$: $\omega = \frac{4\cdot 22}{7\pi} = \frac{88}{7\pi}$ Using $\pi = \frac{22}{7}$, $\omega = \frac{88}{7\cdot \frac{22}{7}} = \frac{88}{22} = 4$

Therefore, $\boxed{4\ \text{rad/s}}$