jmc

Let $s$ be the side length of the square, and let $h$ be the length of the altitude of $△ABC$ from $B$. Because $△ABC$ and $△WBZ$ are similar, it follows that $$\frac{h-s}{s}=\frac{h}{AC}=\frac{h}{5},\text{so}s=\frac{5h}{5+h}.$$Because $h=3\cdot 4/5=12/5$, the side length of the square is $$s=\frac{5(12/5)}{5+12/5}=\boxed{\frac{60}{37}}.$$ OR

Because $△WBZ$ is similar to $△ABC$, we have $$BZ=\frac{4}{5}s\text{and}CZ=4-\frac{4}{5}s.$$Because $△ZYC$ is similar to $△ABC$, we have $$\frac{s}{4-(4/5)s}=\frac{3}{5}.$$Thus $$5s=12-\frac{12}{5}s\text{and}s=\boxed{\frac{60}{37}}.$$