Answer
$K_a=K_b\gt K_c$
Work Step by Step
First of all, we need to find the moment of inertia for each object of them. Recall that the moment of inertia of a disk is given by $\dfrac{1}{2}MR$. We know that the three disks are having the same mass.
Thus,
$$I_a=\dfrac{1}{2}m_aR_a^2=\boxed{\dfrac{1}{2}m r^2}$$
$$I_b=\dfrac{1}{2}m_bR_b^2=\dfrac{1}{2}m \left(\frac{1}{2}r\right)^2=\boxed{\dfrac{1}{8}mr^2}$$
$$I_c=\dfrac{1}{2}m_cR_c^2=\dfrac{1}{2}m \left(2r\right)^2=\boxed{2mr^2}$$
Now we can find their kinetic energies.
$$K_a=\dfrac{1}{2}I_a\omega_a^2=\dfrac{1}{2}\left[\dfrac{1}{2}m r^2\right]\omega^2$$
$$K_a= \dfrac{1}{4} m r^2\omega^2\tag a$$
$$K_b=\dfrac{1}{2}I_b\omega_b^2=\dfrac{1}{2}\left[\dfrac{1}{8}m r^2\right](2\;\omega)^2$$
$$K_b= \dfrac{1}{4} m r^2\omega^2\tag b$$
$$K_c=\dfrac{1}{2}I_c\omega_c^2=\dfrac{1}{2}\left[2m r^2\right]\left(\dfrac{1}{3}\;\omega\right)^2$$
$$K_b= \dfrac{1}{9} m r^2\omega^2\tag c$$
From above, it is obvious that $$\boxed{K_a=K_b\gt K_c}$$
