The spring has a length $l .$ Whem $m$ is placed over it, the equilibrium position becomes $O^{\prime}$

If it is pressed from $O^{\prime}$ $(the\, equilibrium\, position)$ to $O^{\prime \prime}, O^{\prime} O^{\prime \prime}$ is the amplitude.

$O O^{\prime} =\frac{m g}{k}=\frac{2 \times 10}{200}=0.10 \mathrm{m} $

$m g =k x_{0}$

If the restoring force $m A \omega^{2}>m g,$ then the mass will move up with acceleration, detached from the pan.

i.e. $A>\frac{g}{k / m} \Rightarrow A>\frac{20}{200}>0.10 \mathrm{m}$

$The\, amplitude >10 \mathrm{cm}$

i.e. the minimum is just greater than $10 \mathrm{cm}$. (The actual compression will include $x_{0}$ also. But when talking of amplitude, it is always from the equilibrium position with respect to which the mass is oscillating.