The energy of the system is E=5.5 J. This energy is the same at every moment of the oscillation. When the stretch x of the spring is maximum (so, when the stretch is equal to the amplitude: x=A) the velocity of the spring is zero, so all this energy is just elastic potential energy of the spring:
[tex]E= \frac{1}{2}kA^2 [/tex]
From which we find
[tex]A= \sqrt{ \frac{2E}{k} }= \sqrt{ \frac{2 \cdot 5.5 J}{1200 N/m} }=0.096 m [/tex]
