The moment of inertia for a system of point masses is $I = \sum m_i r_i^2$, where $m_i$ is the mass of the i-th particle and $r_i$ is its distance from the axis of rotation. Let the mass of each ball be $m = 0.2$ kg and the rod length be $L = 0.8$ m.

[Q1] The axis of rotation is at the center of the rod. Each ball is at a distance of $r = L/2$ from the axis. The total moment of inertia $I_1$ is the sum of the moments of inertia of the two balls.

$$I_1 = m\left(\frac{L}{2}\right)^2 + m\left(\frac{L}{2}\right)^2 = 2m\frac{L^2}{4} = \frac{1}{2}mL^2$$ $$I_1 = \frac{1}{2}(0.2 \text{ kg})(0.8 \text{ m})^2 = 0.064 \text{ kg m}^2$$

[Q2] The axis of rotation passes through one of the balls. One ball is at a distance $r_1=0$ from the axis, and the other is at a distance $r_2=L$. The total moment of inertia $I_2$ is:

$$I_2 = m(0)^2 + m(L)^2 = mL^2$$ $$I_2 = (0.2 \text{ kg})(0.8 \text{ m})^2 = 0.128 \text{ kg m}^2$$