Answer:
A) at the rod's geometrical center
Explanation:
Let us assume that the rod is replaced by water. And now this water volume is in translational and in rotational equilibrium.
Therefore, a net upward force must have been exerted by the surrounding liquid which acts at the center of mass of the water volume.
This force determines through the geometric center of the column of the cylindrical water
Moreover, the force is also independent of submerged body into it
Hence, the first option is correct
The pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.
Explanation:
In the present problem, the temperature is said to remain at constant and there is change in the pressure. So according to Boyle's law, the relationship between pressure and volume of any gaseous objects are inversely related to each other. In other words, the pressure attained by gas molecules in a container will be inversely proportional to the volume of the gas molecules occupied in the container, at constant temperature.

So, if two volumes V₁ and V₂ are considered, then their respective pressure will be represented as P₁ and P₂. Then, as per Boyle's law,

So let us consider, V₁ = 6 cm³ and V₂ = 4 cm³ and pressure P₁ = 405 kPa and we have to determine P₂.
Then, 
So, the pressure at new volume of 4 cm³ is 486 kPa. It can be seen that as there is decrease in the volume, there is an increase in the pressure. So it satisfied the Boyle's law.
Thus, the pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.
For Ar :
1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L
x = 7.6 * 22.4
x = 170.24 L
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For C2H3:
1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L
y = 0.44 * 22.4
y = 9.856 L
hope this helps !.