A uniform metre rule, 1.00 m long, has a small hole at one end…

(a) The period. $I = \tfrac13 mL^2$ and $d = \tfrac{L}{2}$, so $T = 2\pi\sqrt{\dfrac{2L}{3g}} = 2\pi\sqrt{\dfrac{2 \times 1.00\ \text{m}}{3 \times 9.8\ \text{m/s}^2}} = 1.64\ \text{s}$.

(b) The equivalent length. $\tfrac{2L}{3} = 0.667\ \text{m}$.

(c) The mistake. $2\pi\sqrt{L/g}$ is for a bob with all its mass at the end of the string. The mass of the rule is spread along it, much of it near the nail. So it swings faster than a $1.00\ \text{m}$ simple pendulum.