Answer
(a) $16000N/m$ (b) More than 180 J.
Work Step by Step
(a) The work needed to compress a spring is equal to $$W=\frac{1}{2}k\Delta x^2$$ Solving for $k$ yields $$k=\frac{2W}{\Delta x^2}$$ Substituting known values of $W=180J$ and $\Delta x=0.15m$ yields a spring constant of $$k=\frac{2(180J)}{(0.15m)^2}=16000N/m$$ (b) The new compression distance is equal to $0.15m+0.15m=0.30m$. Substituting known values of $\Delta x=0.30m$ and $k=16000N/m$ yields a work $$W=\frac{1}{2}(16000N/m)(0.30m)^2=720J$$ This change in work is $720J-180J=520J$, which is more than $180$ J.
