b
              According to the theorem of parallel axes, the moment of inertia of the thin rod of mass \(M\) and length \(L\) about an axin passing through one of the ends is

\(I = {I_{CM}} + M{d^2}\)

Where \({I_{CM}}\) is the moment of inertia of the given rod about an axis passing through its center of mass and perpendicular to its lenght and \(d\) is the distance between two parallel axes.

Herer, \({I_{CM}} = {I_{0'}}\,d = \frac{L}{2}\)

\(\therefore \,I = {I_0} + M{\left( {\frac{L}{2}} \right)^2} = {I_0} + \frac{{M{L^2}}}{4}\)