Ruler and compass construction of the unit-distance petersen graph embedding

If the inner star's side are $1$ the radius of its circumflexing circle is $\frac1{2r}=\sin\frac25\pi=\sqrt{\frac{5+\sqrt5}8}$ which leads that $$r=\sqrt{\frac{5-\sqrt5}{10}}$$

So you can build $r$ with ruller and compass.

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Update

Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:

Then draw perpendiculars to each side at each vertex.

These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).

The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)

So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.