It takes 2.5 J of energy to stretch a helical spring by 15 cm. Determine the spring constant.

Past exam question for physics. Any help is appreciated.

The elastic potential energy stored in a spring is:

Ee = k × x² / 2
where
Ee = elastic potential energy = 2.5 J
k = spring constant = ?
x = spring stretching = 15 cm = 0.15 m
so
(2.5 J) = k × (0.15 m)² / 2
k = 222.22 N/m

u=1/2k*x^2
2.5=1/2k*(225/100)
0.5k=1.11......
k=2.22....