The question involves comparing the densities of two liquids based on the floating behavior of an object. To solve it, we'll apply the principle of buoyancy which states that an object will float in a liquid if the weight of the liquid displaced by the submerged part of the object is equal to the weight of the object.

Let's denote the density of the object as \(\rho_o\), the density of liquid A as \(\rho_A\), and the density of liquid B as \(\rho_B\). According to the information given:

• The object floats with \(\frac{1}{4}\)th part of its volume outside the liquid in liquid A. Therefore, \(\frac{3}{4}\)th of its volume is submerged.
• The object floats with \(\frac{3}{4}\)th part of its volume outside the liquid in liquid B. Therefore, \(\frac{1}{4}\)th of its volume is submerged.

From Archimedes' principle:

• The weight of the liquid displaced by the submerged part is equal to the weight of the object.
• When in liquid A, the buoyant force is \(\rho_A \times g \times \frac{3}{4} \, V = \rho_o \times g \times V\), where \(V\) is the volume of the object.
• Simplifying, we find that \(\rho_A \times \frac{3}{4} = \rho_o\).
• When in liquid B, the buoyant force is \(\rho_B \times g \times \frac{1}{4} \, V = \rho_o \times g \times V\).
• Simplifying, we find that \(\rho_B \times \frac{1}{4} = \rho_o\).

By comparing the equations, we have:

• \rho_A \times \frac{3}{4} = \rho_B \times \frac{1}{4}
• This gives us: \frac{\rho_A}{\rho_B} = \frac{1}{3}
• Therefore, \rho_B = 3 \rho_A

This shows that the density of liquid B is greater than the density of liquid A. Thus, the correct answer is:

This analysis is based entirely on the application of Archimedes' principle and the given conditions about how the object floats in two different liquids with different submerged volumes.