An aluminum block weighing 0.54 kg is completely immersed in water, while a buoyant force equal to 2 N

An aluminum block weighing 0.54 kg is completely immersed in water, while a buoyant force equal to 2 N acts on it. What is the weight of the displaced water?

Using the basic formula of the Archimedes force, we find the volume of the submerged part of the body. Then, knowing the density of the body (the density of water and aluminum, look in the density table), we find the volume of the whole body. Then we divide the lesser by the greater (since we need to find out what part of the total volume is loaded).
Condition: m = 0.54 kg; Fn = 2H; gv = 1 * 10 ^ 3; gt = 2.7 * 10 ^ 3
It is necessary to find Vdives / Vall -?
Decision:
Fa = gzh * g * Vlim
Let us express from the formula V
Vlim = Fn / (gzh * g) = 2 / (1 * 10 ^ 3 * 10) = 0.2 * 10 ^ 3 = 200m ^ 3
Vs = m / gt = 0.54 / (2.7 * 10 ^ 3) = 0.2 * 10 ^ 3 = 200m ^ 3
It turns out the same area, but this cannot be. It is possible that there is a typo in the condition and then, with a weight of 5.4 kg, Vw = 2000m ^ 3.
Vlim / Vweight = 200/2000 = 2/20 = 1/10
Answer: 1/10