The period of angular SHM is $T = 2\pi\sqrt{I/\kappa}$. The moment of inertia of a rod of length $L$ rotating about its center is $I = \frac{1}{12}mL^2$.

For a small angular displacement $\theta$, the end of the rod moves a distance $x \approx (L/2)\theta$, stretching the spring. The spring exerts a restoring force $F = kx = k(L/2)\theta$. The force is nearly perpendicular to the rod, with a lever arm of $L/2$. The restoring torque is $\tau = -F(L/2) = -[k(L/2)\theta](L/2) = -k(L/2)^2\theta$. The torsional constant is $\kappa = k(L/2)^2$.

The period of oscillation is:

$$T = 2\pi\sqrt{\frac{I}{\kappa}} = 2\pi\sqrt{\frac{\frac{1}{12}mL^2}{k(L/2)^2}} = 2\pi\sqrt{\frac{mL^2/12}{kL^2/4}} = 2\pi\sqrt{\frac{m}{3k}}$$

The length $L$ cancels out.

$$T = 2\pi\sqrt{\frac{0.600 \text{ kg}}{3(1850 \text{ N/m})}} = 0.0653 \text{ s}$$