Answer
$\frac{K}{K_0} = 0.92$
Work Step by Step
We can find the initial rotational inertia of the system:
$I_0 = \frac{1}{2}mr^2+\frac{1}{2}(10~m)(3.0~r)^2$
$I_0 = \frac{1}{2}mr^2+\frac{1}{2}~(90~m~r^2)$
$I_0 = \frac{91~m~r^2}{2}$
We can find the final rotational inertia of the system:
$I_f = \frac{1}{2}mr^2+m(2r)^2+\frac{1}{2}(10~m)(3.0~r)^2$
$I_f = \frac{9}{2}mr^2+\frac{1}{2}~(90~m~r^2)$
$I_f = \frac{99~m~r^2}{2}$
We can use conservation of angular momentum to find the final angular velocity:
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{\frac{91~m~r^2}{2}~\omega_0}{\frac{99~m~r^2}{2}}$
$\omega_f = \frac{91~\omega_0}{99}$
$\omega_f = \frac{(91)~(20~rad/s)}{99}$
$\omega_f = 18.38~rad/s$
We can find the ratio $\frac{K}{K_0}$:
$\frac{K}{K_0} = \frac{\frac{1}{2}I_f~\omega_f^2}{\frac{1}{2}I_0~\omega_0^2}$
$\frac{K}{K_0} = \frac{I_f~\omega_f^2}{I_0~\omega_0^2}$
$\frac{K}{K_0} = \frac{(\frac{99~m~r^2}{2})~(18.38~rad/s)^2}{(\frac{91~m~r^2}{2})~(20~rad/s)^2}$
$\frac{K}{K_0} = \frac{(99)~(18.38~rad/s)^2}{(91)~(20~rad/s)^2}$
$\frac{K}{K_0} = 0.92$
