A thin uniform rod of mass $M$ and length $L$ rotates about an axis perpendicular to the rod and passing through one of its ends. Its moment of inertia is:

Using the parallel-axis theorem: $I_{\text{end}} = I_{\text{cm}} + M d^2 = \dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{12} + \dfrac{ML^2}{4} = \dfrac{ML^2}{3}$. Option A is the centre-axis value — the classic mistake of forgetting to shift the axis.