Answer
$\omega_f=0.037rad/s$
Work Step by Step
Because angular momentum is conserved, $$I_0\omega_0=I_f\omega_f$$ $$\omega_f=\frac{I_0\omega_0}{I_f}$$
At first, the disk's moment of inertia $I_0=0.1kg.m^2$ and angular velocity $\omega_0=0.067rad/s$
After sand is dropped, $I_f=0.1+M_{sand}R^2=0.1+0.5\times0.4^2=0.18kg.m^2$
Therefore, $\omega_f=0.037rad/s$
