Answer
See answer.
Work Step by Step
Let the block have a cross-sectional area of A. At equilibrium, the net force on the block is zero, so $F_{buoyant}=mg$.
Let the block be pushed downward into the water by an additional distance x. There is additional water displaced, therefore an increase in the upward buoyant force, equal to the weight of the additional water.
Calculate the mass as the density of water multiplied by the displaced volume, which is "Ax". The weight is mg.
$$F_{extra}=m_{extra}g=\rho_{water}Axg=(\rho_{water}Ag)x $$
This is the net force on the displaced block. This restoring force is upward, and proportional to the displacement x. A similar argument shows that when the block is lifted upward by a distance x, the buoyant force decreases by the same amount we just calculated.
In other words, for both upward and downward displacement, there is a net restoring force of magnitude $F_{extra} =(\rho_{water}Ag)x $.This is the condition for simple harmonic motion, with a “spring constant” of $k=(\rho_{water}Ag) $.
