Answer
$\omega=0.26rad/s$
Work Step by Step
When the bug is at the axis, it contributes no moment of inertia to the system. So the total moment of inertia is $$I_0=I_{rod}=1.1\times10^{-3}kg.m^2$$
When the bug is at the other end of the rod, it is $0.25m$ away from the axis, so the total moment of inertia is $$I=I_{rod}+m_{bug}(0.25m)^2=1.36\times10^{-3}kg.m^2$$
The rod's initial velocity $\omega_0=0.32rad/s$. Angular momentum is conserved, so $$I_0\omega_0=I\omega$$ $$\omega=\frac{I_0\omega_0}{I}=0.26rad/s$$
