This is very difficult to explain without free-bodydiagrams (FBD). I'll try to draw something:

The stuff on the right is supposed to be a spring.

The FBD for the top mass:

The bottom mass:

\(\displaystyle{f}=\mu{n}=\mu{m}{g}\)

F is the force due to the spring (F=kx). Applying Newton's 2nd law to the first FBD:

\(\displaystyle\mu{m}{g}={m}{a}\)

And to the second FBD:

\(\displaystyle{k}{x}-\mu{\left({2}{m}\right)}{g}={m}{a}\)

The blocks don't slide when they have the same acceleration:

\(\displaystyle{a}=\mu{g}={\frac{{{k}{x}}}{{{m}}}}-{2}\mu{g}\)

\(\displaystyle\mu={\frac{{{k}{x}}}{{{3}{m}{g}}}}\)

m is the mass of a single block, calculated in the first partof the problem. The blocks will slide when x = 0.05; plugthis value for x into the equation and solve.

The stuff on the right is supposed to be a spring.

The FBD for the top mass:

The bottom mass:

\(\displaystyle{f}=\mu{n}=\mu{m}{g}\)

F is the force due to the spring (F=kx). Applying Newton's 2nd law to the first FBD:

\(\displaystyle\mu{m}{g}={m}{a}\)

And to the second FBD:

\(\displaystyle{k}{x}-\mu{\left({2}{m}\right)}{g}={m}{a}\)

The blocks don't slide when they have the same acceleration:

\(\displaystyle{a}=\mu{g}={\frac{{{k}{x}}}{{{m}}}}-{2}\mu{g}\)

\(\displaystyle\mu={\frac{{{k}{x}}}{{{3}{m}{g}}}}\)

m is the mass of a single block, calculated in the first partof the problem. The blocks will slide when x = 0.05; plugthis value for x into the equation and solve.