Answer
Natural period ($\tau$) is $2\pi/\sqrt {(\frac{3k}{m}+\frac{3g}{2L})}$
Work Step by Step
By summing the moments about point $A$:
$\circlearrowleft +\Sigma M_{A}=I_{A}\ddot \theta$, where $\theta$ is the rotation angle of the rod around $A$
Because the rod is uniform $I_{A}=\frac{mL^{2}}{3}$
$\circlearrowleft +\Sigma M_{A}=-(kLsin\theta)\times L-(mgsin\theta)\times \frac{L}{2}=\frac{mL^{2}}{3}\ddot \theta...eq(1)$
Since the sideward displacement is small $sin\theta=\theta$
Rewriting $eq(1)$:
$\ddot \theta+(\frac{3k}{m}+\frac{3g}{2L})\theta=0$ $\Rightarrow$ $\ddot \theta+\omega_{n}^{2}\theta=0$
$\Rightarrow$ $\omega_{n}=\sqrt {(\frac{3k}{m}+\frac{3g}{2L})}$
To determine the natural period $(\tau)$:
$\tau=\frac{2\pi}{\omega_{n}}$
$\tau=2\pi/\sqrt {(\frac{3k}{m}+\frac{3g}{2L})}$
