ANSWER:
189.52 kg*m^2
Explanation:
Given:
mr = 8 kg
ms = 36.25 kg
L = 6 m
R = 1.5 m
Moment of inertia of rod about its center of mass:
![I_{\operatorname{cm}}=(1)/(12)\cdot m_r\cdot L^2]()
Parallel axis theorem for the rod gives:

Paraller axis theorem for the spehere gives:

Therefore:
![\begin{gathered} I=I_{\operatorname{cm}}+I_r+I_s \\ I=(1)/(12)m_rL^2+m_r((L)/(2)+(R)/(2))^2+(13)/(20)m_sR^2 \end{gathered}]()
Replacing:

The moment of inertia is 189.52 kg*m^2