Answer
$0.58 \;\rm s$
Work Step by Step
This is a physical pendulum and we need to find its equation.
We assume that $\theta$, as shown in the figure below, is a small angle.
Now we need to find the net torque exerted on our rod around the pivot point.
We chose counterclockwise to be the positive torque direction.
From the geometry of the right tiangle below.
$$\sum\tau=I\alpha=-F_{sp}y-mg d$$
the spring force is given by $kx$
$$ I\alpha=-kxy-mg y$$
where $x$ is given by $\sin\theta=x/L$, so $x=L\sin\theta$,
$y$ is given by $\cos\theta=y/L$, so $y=L\cos\theta$,
and $d$ is given by $\sin\theta=d/(L/2)$, so $d=L\sin\theta/2$
$$ I\alpha=-kL^2\sin \theta\cos\theta-\frac{1}{2}mgL\sin\theta$$
Recall that the moment of inertia of the rod at one end is $I=mL^2/3$, and that $\alpha=d^2\theta/dt^2$
$$\frac{1}{3}mL^2\dfrac{d^2\theta}{dt^2}=-kL^2\sin \theta\cos\theta-\frac{1}{2}mgL\sin\theta$$
Dividing both sides by $\frac{1}{3}mL^2$
$$ \dfrac{d^2\theta}{dt^2}=-\dfrac{3k}{m} \sin \theta\cos\theta-\frac{3g}{2L} \sin\theta$$
Recall that $\sin(2\theta)=2\sin\theta \cos\theta$, so $\sin\theta\cos\theta=\dfrac{1}{2}\sin(2\theta)$
$$ \dfrac{d^2\theta}{dt^2}=-\dfrac{3k}{2m}\sin(2\theta)-\frac{3g}{2L} \sin\theta$$
We know, for small angles, that $\sin\theta\approx \theta$ and $\sin(2\theta)\approx 2\theta$
$$ \dfrac{d^2\theta}{dt^2}=-\dfrac{3k}{\color{red}{\bf\not} 2m}(\color{red}{\bf\not} 2\theta)-\frac{3g}{2L} (\theta)$$
$$ \dfrac{d^2\theta}{dt^2}=-\left[\dfrac{3k}{m}+\frac{3g}{2L} \right]\theta$$
This formula is similar to
$\dfrac{d^2x}{dt^2}=-\dfrac{k}{m}x$ and $\dfrac{d^2s}{dt^2}=-\dfrac{g}{L}s$ and both are for simple harmonic motion.
Therefore,
$$\omega=\sqrt{\dfrac{3k}{m}+\frac{3g}{2L}}$$
where $\omega=2\pi f$ and $f=1/T$, so
$$ \dfrac{2\pi}{T}=\sqrt{\dfrac{3k}{m}+\frac{3g}{2L}}$$
$$T=\dfrac{2\pi}{\sqrt{\dfrac{3k}{m}+\dfrac{3g}{2L}}}$$
Plugging the known;
$$T=\dfrac{2\pi}{\sqrt{\dfrac{3(3)}{0.200}+\dfrac{3(9.8)}{2(0.2)}}}=0.577\;\rm s$$
$$T=\color{red}{\bf 0.58}\;\rm s$$
