What is the Archimedean force that acts on a body dropped into a measuring glass of water?

What is the Archimedean force that acts on a body dropped into a measuring glass of water? Before lowering the body into a glass, there were 200 ml of liquid in the beaker, after it was lowered, it became 250 ml.

Given:
V1 = 200 milliliters = 0.0002 m ^ 3 – the volume of water in the beaker;
V2 = 250 milliliters = 0.00025 m ^ 3 – the volume of water in the beaker after the body was lowered into it;
ro = 1000 kg / m ^ 3 – water density;
g = 9.8 Newton / kilogram – acceleration of gravity.
It is required to determine A (Newton) – Archimedean force acting on the body.
Let’s find the volume of the body, which is lowered into the beaker:
V = V2 – V1 = 0.00025 – 0.0002 = 0.00005 m ^ 3.
Then the Archimedean force will be equal to:
A = ro * V * g = 1000 * 0.00005 * 9.8 = 0.5 Newton.
Answer: The Archimedean force is 0.5 Newton.