Verified answer

I am not sure what you mean by 'when it is *****'...I am assuming you mean when it is vertical

we solve this using conservation of energy

when the spring is compressed, it has PE equal to 1/2 k x^2 where k is the spring constant and x is the displacement from equilibrium

when it is at its maximum height, it has PE of m g H where H is the height above its launch point, so we have

1/2 k x^2 = m g H

H = k x^2/2 m g = 300N/m *(0.05m)^2/2*0.05kg*9.8m/s/s = 0.77m

this is the height above the initial launch point; if the question is asking how high it gets above the equilibrium position, remember to subtract 0.05m since the spring was depressed by that amount