An iron piece measuring 20 * 20 * 25 cm is immersed in water. What force must be applied to keep this piece in water?

An iron piece measuring 20 * 20 * 25 cm is immersed in water. What force must be applied to keep this piece in water? The density of water is 1000 kg / m, the density of iron is 7800 kg / m.

a = 20 cm = 0.2 m.
b = 20 cm = 0.2 m.
h = 25 cm = 0.25 m.
ρw = 1000 kg / m3.
ρzh = 7800 kg / m3.
g = 9.8 m / s2.
F -?
A body immersed in a liquid or gas is affected by 2 forces: the force of gravity Ft and the force of Archimedes Farkh.
The force of gravity Ft, directed vertically downward and is determined by the formula: Ft = m * g = V * ρzh * g.
The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρw * g * V. Where ρw is the density of the liquid in which the body is immersed, g is the acceleration of gravity, V is the volume of the immersed part of the body in the liquid.
F = Ft – Farch = V * ρl * g – V * ρw * g = V * g * (ρl – ρw).
Since the part has the shape of a rectangular parallelepiped, its volume V is expressed by the formula: V = a * b * h.
F = a * b * h * g * (ρzh – ρv).
F = 0.2 m * 0.2 m * 0.25 m * 9.8 m / s2 * (7800 kg / m3 – 1000 kg / m3) = 666.4 N.
Answer: to keep the part under water, it is necessary to apply a force F = 666.4 N directed vertically upward to it.