The period $T$ of a torsional pendulum is given by $T = 2\pi \sqrt{I/\kappa}$, where $I$ is the rotational inertia and $\kappa$ is the torsion constant of the wire.

For the rod, the rotational inertia about its midpoint is $I_a = \frac{1}{12}mL^2$. The period is:

$$T_a = 2\pi \sqrt{\frac{I_a}{\kappa}}$$

For object X with rotational inertia $I_X$, the period is:

$$T_b = 2\pi \sqrt{\frac{I_X}{\kappa}}$$

The torsion constant $\kappa$ is the same for both setups. Dividing the second equation by the first gives:

$$\frac{T_b}{T_a} = \frac{2\pi \sqrt{I_X/\kappa}}{2\pi \sqrt{I_a/\kappa}} = \sqrt{\frac{I_X}{I_a}}$$

Solving for $I_X$:

$$I_X = I_a \left(\frac{T_b}{T_a}\right)^2$$

Substituting the expression for the rod's rotational inertia, $I_a$:

$$I_X = \frac{1}{12}mL^2 \left(\frac{T_b}{T_a}\right)^2$$