Solution

Let the bar be rotated through a small angle $$\theta$$.
The restoring torque of the forces $$mg, k_1 x$$ and $$k_2 x$$ about $$O$$ can be given as
$$\tau =-\left[mg \left(\dfrac 12 \right)\sin \theta +k_1 x (l \cos \theta)+k_2 x(l \cos \theta)\right]$$
Since $$\theta $$ is small $$\sin \theta \cong \theta , x=l\ \theta$$ and $$\cos \theta \cong 1$$
Putting $$k_1 +k_2$$, we obtain
$$\tau =-\left[kl^2 +mg\left(\dfrac 12 \right)\right] \theta$$
or $$I \alpha =-\left[kI^2 +mg \left(\dfrac 12\right)\right]\theta$$
$$\Rightarrow \ \omega_{osc}=\sqrt {\left(\dfrac {kl^2+mg(l/2)}{(ml^2 /3)}\right)} =\sqrt {\dfrac {3k}{m}+\dfrac {3g}{2l}}$$