Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. The amplitude of oscillations is 

Initially, the spring supports both masses, so the equilibrium extension is $$x_1=\frac{(m_1+m_2)g}{K}.$$ After removing $m_1$, the new equilibrium extension for the remaining mass is $$x_2=\frac{m_2g}{K}.$$ At the instant $m_1$ is removed, $m_2$ is still at the old equilibrium position and has zero velocity, so its displacement from the new equilibrium position is $$A=x_1-x_2.$$ Therefore, $$A=\frac{(m_1+m_2)g}{K}-\frac{m_2g}{K}=\frac{m_1g}{K}.$$ This matches option $A$.