Answer
Time period of physical pendulum (T)=
$$2π\sqrt \frac{I}{mgd}$$
Where $I$=Moment of inertia of body at point of suspension, $m$=mass of body, g=gravitational acceleration, $d$=distance between center of mass and point of suspension
Work Step by Step
Let $l$ be length of rod, then
moment of inertia ($I$) at point of suspension = $\frac{1}{3}ml^{2}$
And $d=\frac{l}{2}$
By Formula,
Time period of physical pendulum (T)=
$2π\sqrt \frac{I}{mgd}$
Puting values of $I$ and $d$,
$T=\sqrt \frac{l}{6g}$
By puting values,
$1.5=2π\sqrt\frac{l}{60}$
By solving
$l=3.423$
