jmc

Squaring the equation $|2z-w|=25$, we get $|2z-w{|}^2=625$. Since $k\cdot \overline{k}=|k{|}^2$ for all complex numbers $k$, we have that $$(2z-w)(2\overline{z}-\overline{w})=625.$$Expanding, we get $$4z\overline{z}-2(w\overline{z}+\overline{w}z)+w\overline{w}=625.$$Similarly, from the equation $|z+2w|=5$, we get $$(z+2w)(\overline{z}+2\overline{w})=25.$$Expanding, we get $$z\overline{z}+2(w\overline{z}+\overline{w}z)+4w\overline{w}=25.$$Finally, from the equation $|z+w|=2$, we get $$(z+w)(\overline{z}+\overline{w})=4.$$Expanding, we get $$z\overline{z}+(w\overline{z}+\overline{w}z)+w\overline{w}=4.$$We then have the equations $$\begin{array}{rl}4z\overline{z}-2(w\overline{z}+\overline{w}z)+w\overline{w}& =625,\\ z\overline{z}+2(w\overline{z}+\overline{w}z)+4w\overline{w}& =25,\\ z\overline{z}+(w\overline{z}+\overline{w}z)+w\overline{w}& =4.\end{array}$$Let $a=z\overline{z}$, $b=w\overline{z}+\overline{w}z$, and $c=w\overline{w}$. Then our equations become $$\begin{array}{rl}4a-2b+c& =625,\\ a+2b+4c& =25,\\ a+b+c& =4.\end{array}$$Adding the first two equations, we get $5a+5c=650$, so $a+c=130$. Substituting into the equation $a+b+c=4$, we get $b+130=4$, so $b=-126$.

Substituting this value of $b$ into the first two equations, we get $4a+252+c=625$ and $a-252+4c=25$, so $$\begin{array}{rl}4a+c& =373,\\ a+4c& =277.\end{array}$$Multiplying the first equation by 4, we get $16a+4c=1492.$ Subtracting the equation $a+4c=277,$ we get $15a=1215$, so $a=81$.

But $a=z\overline{z}=|z{|}^2$, so $|z|=\boxed{9}$.