Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 325: 60a

$I = 0.24~kg~m^2$

Let $m_B$ be the mass of the block. Let $m_b$ be the mass of the bullet. We can find the rotational inertia of the system: $I = (0.060~kg~m^2)+m_B~R^2+m_b~R^2$ $I = (0.060~kg~m^2)+(0.50~kg)(0.60~m)^2+(0.0010~kg)(0.60~m)^2$ $I = 0.24~kg~m^2$