Equilibrium and small-displacement analysis. The rod is hinged at the bottom; tilting it by a small angle $\theta$ from vertical changes the wetted volumes in the two layers and creates a restoring torque from buoyancy.

Buoyant forces on the two halves. The lower half (length $L/2$) is in the heavier liquid, contributing $F_{B1}=(6\rho )(AL/2)g=3\rho ALg$, acting at $L/4$ from the hinge. The upper half is in the lighter liquid, contributing $F_{B2}=(2\rho )(AL/2)g=\rho ALg$, acting at $3L/4$ from the hinge.

Weight of the rod. $W=\rho ALg$, acting at $L/2$ from the hinge.

Restoring torque (for small angular displacement $\theta$). The horizontal lever arms of the gravity and buoyancy forces produce a net torque:

$\tau =\sin \theta \left[F_{B1}\cdot \frac{L}{4}+F_{B2}\cdot \frac{3L}{4}-W\cdot \frac{L}{2}\right]$

$=\sin \theta \cdot \rho A{L}^{2}g\left[\frac{3}{4}+\frac{3}{4}-\frac{1}{2}\right]=\rho A{L}^{2}g\cdot \theta$ (taking $\sin \theta \approx \theta$).

With the sign convention this is restoring, so $\tau =-\rho A{L}^{2}g\theta$.

Moment of inertia about the hinge: $I=\frac{1}{3}m{L}^2=\frac{1}{3}\rho AL\cdot L^2=\frac{\rho A{L}^3}{3}$.

Angular SHM: $\ddot{\theta }=-\frac{\rho A{L}^{2}g}{\rho A{L}^3/3}\theta =-\frac{3g}{L}\theta$, so $T=2\pi \sqrt{L/(3g)}=\frac{2\pi }{\sqrt{3}}\sqrt{L/g}$.

$\therefore n=\sqrt{3}\approx 1.73$.