#### Answer

(a) $I = 3.8\times 10^{-5}~kg~m^2$
(b) $I = 1.1\times 10^{-4}~kg~m^2$

#### Work Step by Step

(a) We can find the moment of inertia of the CD for a perpendicular axis through the center.
$I = \frac{1}{2}MR^2$
$I = \frac{1}{2}(0.021~kg)(0.060~m)^2$
$I = 3.8\times 10^{-5}~kg~m^2$
(b) We can use the parallel axis theorem to find the moment of inertia about a perpendicular axis that is through the edge of the disk. Note that the distance $d$ from the center of mass to the edge of the disk is 6.0 cm.
$I = I_{cm}+Md^2$
$I = 3.8\times 10^{-5}~kg~m^2+(0.021~kg)(0.060~m)^2$
$I = 1.1\times 10^{-4}~kg~m^2$