Answer
$ \frac{1}{2}kx_0^2 = \mu mgd$
Work Step by Step
We know that the block will stop when all of the spring energy becomes mechanical energy, meaning that it will be lost to friction. Thus, we find:
$ \frac{1}{2}kx_0^2 = F_f \times d$
$ \frac{1}{2}kx_0^2 = \mu F_n \times d$
$ \frac{1}{2}kx_0^2 = \mu mgd$
