Answer
a) $I=2.67kg.m^2$
b) The distance is $1.16m$.
Work Step by Step
a) The moment of inertia of a rod relative to an axis perpendicular to the rod at one end is $$I=\frac{1}{3}MR^2=\frac{1}{3}(2kg)(2m)^2=2.67kg.m^2$$
b) When the mass of the rod is located at a single point that is at a distance $x$ from the axis, the moment of inertia of the system is $$I=Mx^2$$
For this moment of inertia to be equal to $2.67kg.m^2$, $$x=\sqrt{\frac{2.67}{M}}=1.16m$$
