A spring having a spring constant of 180 N/m is stretched 0.51 m from its equilibrium position. How much elastic potential energy does the spring possess?

I'm assuming that you are asking what is the elastic
potential energy stored in the spring at the position
stretched by 16.5 cm.
Since you know the spring constant k, 144 N/m and
the spring stretch from the equilibrium position x, is
16.5 cm, or 0.165, you find the spring's potential
1
energy from the equation PE
-ka?, which equals
2
1.96 Joules, or kg * m?/s? if you want SI units.

Divide 108.2 x 10^6 into 1.5 x 10 and the result will be in au (astronomical units)