Answer
Less than.
Work Step by Step
We know that the angular momentum is given by
$$L=I\omega$$
Thus,
$$\dfrac{L_a}{L_b}=\dfrac{I_a\omega_a}{I_b\omega_b}$$
Recalling that the moment of inertia of a disk is given by $$I=\frac{1}{2}mR^2$$
Thus,
$$\dfrac{L_a}{L_b}=\dfrac{\color{red}{\bf\not} \frac{1}{2}m_aR_a^2\omega_a}{\color{red}{\bf\not} \frac{1}{2}m_bR_b^2\omega_b}$$
Plugging from the data in the given figure.
$$\dfrac{L_a}{L_b}=\dfrac{ \color{red}{\bf\not} m r_a^2\omega_a}{ \color{red}{\bf\not} m(2r_a)^2(\frac{1}{2}\omega_a)}=\dfrac{\color{red}{\bf\not} r_a^2\color{red}{\bf\not} \omega_a}{2\color{red}{\bf\not} r_a^2\color{red}{\bf\not} \omega_a}$$
Hence,
$$\boxed{L_b=2L_a}$$
Therefore, the angular momentum of disk a is less than the angular momentum of disk b.
