A uniform lamina of mass m is bounded by concentric circles with centre O and radii a and 2a. the lamina is free to rotate about a fixed smooth horizontal axis T which is tangential to the outer rim (see diagram). Show that the moment of inertia of the lamina about T is $\frac{21}4ma^2$.
I started with the fact that the moment of inertia of a disk through the centre of mass but not perpendicular to the surface is $\frac14mr^2$ and using parallel axis theorem I obtained the correct equation:
$$I_T = a^2(m_{2a}-\frac14m_a) + 4ma^2$$
Where m is known as $m=m_{2a}-m_a$
But I couldn't figure out how to calculate the values of $m_{2a}$ and $m_a$ in terms of $m$ to get the required answer.