Answer
the ratio $\frac{L_{C}}{L_{B}}=\frac{2^{4}}{(1/4)^{3}}=2^{10}=1024$.
Work Step by Step
Because the disk is connected by a belt, the linear speed of the rim of the disk will be the same,
For disks A and C, $\omega_{C}2R=\omega_{A}R$, or $\omega_{A}=2\omega_{C}$.
For disks A and B, $\omega_{A}R/2=\omega_{B}R/4$, or $\omega_{A}=\omega_{B}/2$.
From the two relations above, we know that $\omega_{B}=4\omega_{C}$
To determine the angular momentum, we need to know the mass of the disk which is $m=\rho \times V=\rho \pi r^{2}h$, with $\rho$ is density; $V$ is volume; $r$ is radius; and $h$ is the thickness of the disk.
Angular momentum of disk C:
$L_{C}=I_{C}\omega_{C}=\frac{1}{2}m_{C}r_{C}^{2}\omega_{C}=\frac{1}{2}\rho \pi r_{C}^{2}hr_{C}^{2}\omega_{C}=\frac{1}{2}\rho \pi h (2R)^{4}\omega_{C}$
Angular momentum of disk B:
$L_{B}=I_{B}\omega_{B}=\frac{1}{2}m_{B}r_{B}^{2}\omega_{B}=\frac{1}{2}\rho \pi r_{B}^{2}hr_{B}^{2}\omega_{B}=\frac{1}{2}\rho \pi h (R/4)^{4}4\omega_{C}$
Thus, the ratio $\frac{L_{C}}{L_{B}}=\frac{2^{4}}{(1/4)^{3}}=2^{10}=1024$.
