A weird arrangement .

I will call the length of the rod L and the mass M

Since the angle is small, the torque would be approximately $\frac{-1}{4}L^2k\theta$ and the rotational inertia is $\frac{ML^2}{12}$ therefore the angular acceleration is approximately $\frac{-3k}{M}\theta$ which gives you the following differential equation:

$\frac{d^2\theta}{dt^2}=\frac{-3k}{M}\theta$

This is the simple harmoic motion differential equation and it's solution is:

$\theta=\phi sin(\sqrt{\frac{3k}{M}}t)$ Where $\phi$ is the maximum angular displacement

The period of the sine function is 2pi, therefore the period of the oscillation is the T that satisfies $\sqrt{\frac{3k}{M}}T=2\pi$

So solving for the period, T, you get:

$T=2\pi\sqrt{\frac{M}{3k}}$

Suppose $r$ is half the length of the rod. Then if $F$ is the force that is acting upon it due to the spring

$F \cdot r = I \ddot{\theta}$

Where $I$ is the moment of inertia of the rod and $\theta$ is the angle the rod makes with the wall. For small $\theta$ one is allowed to write

$\theta \approx \sin \theta = \frac{\Delta x}{r} \;\;\; \therefore \;\;\; \Delta x = r \theta$

Which means that $F = -k \Delta x \approx -kr \theta$ . Therefore, according to the first equation

$-kr^2 \theta \approx I \ddot{\theta} = \frac{mr^2}{3} \ddot{\theta}$

Where we have used $I = mr^2/3$ . Therefore

$\ddot{\theta} + \frac{3k}{m} \theta \approx 0 \;\;\; \Rightarrow \;\;\; T \approx 2\pi \sqrt{\frac{m}{3k}} \approx 0.0653 s$

Let E be the energy of the system which remains conserved. When one end of the rod has undergone a small displacement $'x'$ [Where $x= \frac{l \theta}{2}$ ] .

E = $\frac{I\omega^2}{2} + \frac{kx^2}{2}$

Differentiating both sides

$\frac{I}{2}\frac{d\omega^2}{dt} + \frac{k}{2}\frac{dx^2}{dt} = 0$

$\frac{I}{2}2\omega\frac{d\omega}{dt} + \frac{k}{2}2x\frac{dx}{dt} = 0$

$I\frac{2v}{l}\frac{d^2\theta}{dt^2} + kxv = 0$

$\frac{d^2\theta}{dt^2} + \frac{kl^2\theta}{4I} = 0$

Comparing with equation of differentiability

$\omega = \sqrt{\frac{kl^2}{4I}} = \sqrt{\frac{3k}{m}}$

$T = 2\pi\sqrt{\frac{m}{3k}} \approx 0.0653s$

Time period will be: $T=2π\sqrt{\frac{m}{3k}}$