Answer
$L = 0.54~m$
Work Step by Step
We can use the following equation for the period of a physical pendulum:
$T = 2\pi~\sqrt{\frac{I}{mgd}}$
Here,
$I$ is the moment of inertia of the object
$d$ is the distance from the center of mass to the pivot
Therefore;
$T = 2\pi~\sqrt{\frac{I}{mgd}}$
$T = 2\pi~\sqrt{\frac{\frac{1}{3}mL^2}{mg~(\frac{L}{2})}}$
$T = 2\pi~\sqrt{\frac{2L}{3g}}$
$L = \frac{T^2~(3g)}{2~(2\pi)^2}$
$L = \frac{(1.2~s)^2~(3)(9.80~m/s^2)}{2~(2\pi)^2}$
$L = 0.54~m$
