Answer
$1.83\;s$
Work Step by Step
For a physical pendulum, time period is given by
$T=2\pi \sqrt {\frac{I}{Mgh}}$
Here,
$I=I_1+I_2$
or, $I=\frac{1}{12}mL^2+\frac{1}{3}mL^2=\frac{5}{12}mL^2$
and $h=\frac{1}{2}\times\frac{L}{2}=\frac{L}{4}$
If mass of each rod is $m$, then total mass, $M=2m$
$T=2\pi \sqrt {\frac{\frac{5}{12}mL^2}{2mg\frac{L}{4}}}$
$T=2\pi \sqrt {\frac{5L}{6g}}$
Substituting the given values
$T=2\pi \sqrt {\frac{5\times 1}{6\times9.81}}\;s$
or, $T=1.83\;s$
Therefore, the pendulum’s period of oscillation about a pin inserted through point A at the center of the horizontal stick is $1.83\;s$
