It is a physical pendulum, and its time period $T=2\pi \sqrt{\frac{I}{mgx}}$

here I = moment of inertia of the pendulum about point of suspension

$I=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$

and x = distance of point of suspension from its COM.

x = L/2

$\Rightarrow T=2\pi \sqrt{\frac{M{L}^2/3}{mgL/2}}=2\pi \sqrt{\frac{2L}{3g}}$

$\Rightarrow T=2\pi \sqrt{\frac{2\times 2}{3\times 10}}=2.3\text{ sec}\text{.}$