= Solution
Separate a ball's orbital motion from its spin. For a <rolling ball between concentric races>, the ball centre lies $2a$ from the axle axis. Let its tangential centre <velocity> be $v$ in the direction of axle motion and let its signed spin <angular velocity> be $\Omega$, positive in the same sense as $\omega$. <Rolling without slipping> at the moving inner contact and stationary outer contact requires
$$
v-a\Omega=a\omega,\qquad v+a\Omega=0.
$$
Hence
$$
\boxed{v=\frac{a\omega}{2},\qquad \Omega=-\frac\omega2,\qquad \omega_{\rm orbit}=\frac{v}{2a}=\frac\omega4.}
$$
The spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total <angular momentum> about the axle is the sum of its orbital <angular momentum> and its <angular momentum about the centre of mass>:
$$
L_z=(2a)mv+I_{\rm cm}\Omega=ma^2\omega-\frac12I_{\rm cm}\omega.
$$
For a uniform solid spherical ball, $I_{\rm cm}=2ma^2/5$, so
$$
\boxed{\mathbf L=\frac45ma^2\boldsymbol\omega.}
$$
The solid-ball assumption supplies the usual <moment of inertia>; the preceding expression in terms of $I_{\rm cm}$ also states the answer for a different internal <mass distribution>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2016/paper-2-bearing-kinematics.png]
{title=Original contact-velocity construction separating orbital motion and opposite ball spin}
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