Answer
1.4 rev/s.
Work Step by Step
There is no net external torque on the wheel-clay system, so its net angular momentum is conserved.
We assume the clay has no initial angular momentum. The clay sticks, so the final angular velocity is the same for both objects.
$$L_{i}=L_{f}$$
$$I_i \omega_i = I_f \omega_f$$
$$\omega_f =\omega_i \frac{I_i }{I_f} $$
$$f_f =f_i \frac{I_i }{I_f} $$
Find initial and final moments of inertia.
$$I_i=\frac{1}{2}M_{wheel}R_{wheel}^2$$
$$I_f=\frac{1}{2}M_{wheel}R_{wheel}^2+\frac{1}{2}M_{clay}R_{clay}^2$$
The initial frequency is given, so evaluate to find the final frequency.
$$f_f =(1.5rev/s) \frac{I_i }{I_f}=1.4 rev/s $$