So, I'm going to make ASCII diagrams to explain this:
this is the starting diagram:
l
l
l _____
l k l l
l/\/\/\/\/\/\/l m l
L________l_____l______________

0
We have a spring k and a mass m in equilibrium.
now, we compress the spring by 0.100 m:
l
l
l _____
l k l l
l/\/\/\/\/l m l
L______l___ _l______________
 
0.100 0
The mass is still at rest, but the spring now has potential energy
U = (1/2) k x^2
now, we release the system:
l
l
l _____
l k l l
l/\/\/\/\/\/\/l m l>v
L________l_____l______________

0
The spring extends until it reaches equilibrium, transferring all of it's potential energy into kinetic energy for the mass.
After this point, the spring and the mass separate, and friction starts to play a role.
Since W = integral (F ds) and F is constant, we can say that W = Fs.
F friction = F Normal * u = mgu, thus
W = mgus
Now since W = dE and dE = U (this line is just saying that change in work is change in energy, and the energy goes from U to 0)
W = U.
substituting in:
mgus = (1/2)kx^2
isolating u:
u = (kx^2)/(2mgs)
then we plug in numbers and constants:
k = 105 N/m
x = 0.100 m
m = 2.00 kg
s = 0.250 m
g = 9.80 m/s
you should get u = 0.107. It should be positive, because the only negative term is squared.