= Solution
The string's no-slip condition gives $v=R\omega$. Conservation of <mechanical energy> from release to the stated height drop yields
$$
mgh=\frac12mv^2+\frac12I\frac{v^2}{R^2}.
$$
Solving for the spool's <moment of inertia>,
$$
\boxed{I=\frac{2mghR^2}{v^2}-mR^2,\qquad A=2mghR^2,\quad B=mR^2.}
$$
The axle does no work because its point of application is fixed, and the ideal string and axle introduce no losses. The spool <mass> $M$ is not needed to infer $I$ from this energy measurement.
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