A pendulum of long thin rod.

Total torque acting on the rod about pivot must be equal to $I_{p}\alpha$ .

$\implies$ $mgxsin\theta =I_{p}\alpha$

Since this is a restoring torque so $-mgxsin\theta =I_{p}\alpha$

$\implies$ $\alpha = \frac{-mgxsin\theta}{I_{p}}$

Since for very small angular displacment $sin\theta\approx\theta$

So, $\alpha = \frac{-mgx\theta}{I_{p}}$

This is an equation of SHM whose angular frequency $w = \sqrt{\frac{mgx}{I_{p}}}$

Therefore its time period $T = \frac{2\pi}{w} = 2\pi \sqrt{\frac{I_{p}}{mgx}}$

$= 2\pi \sqrt{\frac{\frac{mL^2}{12} + {mx^2}}{mgx}}$

$= 2\pi\sqrt{\frac{L^2}{12gx} + \frac{x}{g}}$

Since A.M. is greater than or equal to G.M. and equality holds if and only if all the terms are equal. So for time period to be minimum $\frac{L^2}{12gx} = \frac{x}{g}$

$\implies$ $x = \boxed{\frac{L}{2\sqrt{3}}}$

Now putting the value of $x$ in $T$ we would get $T = 2\pi\sqrt{\frac{L}{\sqrt{3}g}}$ .

Therefore $\boxed{T = 2.262 seconds}$ .

Why is it even lvl 4 highly overrated ,but still

The distance required comes out to be: x=l/(2sqrt(3)); plugging, we get T=2pisqrt(l/g*sqrt(3)) which approximately equals:2.2392 s