Answer
$415N/m$
Work Step by Step
We can find the force constant of the spring as follows:
$W=E_f-E_i$
$W=(K.E_f+U_f)-(K.E_i+U_i)$
$\implies W=(0+\frac{1}{2}Kx^2)-(\frac{1}{2}mv^2+0)$
We know taht
$W=-f_kd=-\mu_kgx$
$\implies -\mu_kmgx=\frac{1}{2}Kx^2-\frac{1}{2}mv_i^2$
This can be rearranged as:
$K=\frac{mv_f^2-2\mu_kmgx}{x^2}$
$\implies K=\frac{m(v_i^2-2\mu_kgx)}{x^2}$
We plug in the known values to obtain:
$K=\frac{(1.80)[(2.00)^2-2(0.560)(9.81)(0.110)]}{(0.110)^2}$
$K=415N/m$
