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**Hint:**Archimedes principle says that whenever a body is partially or completely merged into a liquid then the body suffers a loss of weight which is equal to the weight of fluid displaced by the body. The weight of the fluid displaced is equal to the buoyancy force which acts on the body in the upwards direction on the body which is merged in the liquid partially or completely.

**Complete step by step solution:**

It is given in the problem that a penguin floats in a fluid of density ${\rho _o}$ and then floats in the density of $0 \cdot 95{\rho _o}$ and after then it floats in the fluid of density $1 \cdot 1{\rho _o}$ and we need to arrange these densities in ascending order of the buoyant force on the penguin and also we need to arrange the fluid densities on the bases of fluid displaced by the penguin in different fluids.

a) As the buoyancy force is equal to the weight of the fluid displaced by the penguin which is always equal, therefore the buoyancy force is equal in all the cases.

b) The volume of the liquid displaced by the body is equal to the weight of the body floating in the liquid.

The weight of the buoyancy force is equal to,

$ \Rightarrow B = \rho gV$

Where $B$ is the buoyancy force, $\rho $ is density, g is the acceleration due to gravity and V is the volume of block till which the block is submerged.

Here as the density changes the volume merged in the fluid changes because the buoyancy force is constant so as the density increases the volume displaced by the penguin will decrease.

So the arrangement in ascending order of the density according to the volume displaced is equal to,

$0 \cdot 95{\rho _o} > {\rho _o} > 1 \cdot 1{\rho _o}$

**The ascending order of densities of the fluid displaced by the penguin is $0 \cdot 95{\rho _o} > {\rho _o} > 1 \cdot 1{\rho _o}$.**

**Note:**Whenever a body is merged in any fluid then an upward force is applied on the body which results in decreasing the weight of the body when the body is completely merged or partially merged in the fluid.

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