[Paul Copioli]

No. It is NOT over 2L. Natural length has nothing to do with elastic potential energy. The only thing that matters is delta x from the zero energy point, which is the natural length. Notice that your spreadsheet never actually calculates Lamda (spring constant) and you never use the 30 (natural length)in your calculation. If you did use the 30 you would get a spring constant of:

2 Joules (or N-m) = (Lamda * .12m * .12m)/(2*.3m), which yields 83.333N,

which is not the correct spring constant and not even the right units for spring constant. The correct calculation is:

2 Joules (or N-m) = (Lamda * .12m * .12m)/(2), which yields 277.778 N/m for spring constant. This yields exactly 25.24 I just rounded for simplification.

Remember, there is no work done at natural length ... none, zero, nada.

Work is the integral of force per unit length (for linear motion). The force of a linear spring is defined as K*x. The integral over x of K*x is 1/2K*x^2. Remember, that in this definition x is defined as zero at the natural length.

I am 100% confident that my calculation is exactly correct given the problem's boundary conditions.