smc

Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$. Let the point of tangency with the inscribed circle and the right larger circles be $T$. Then $OT=BO+BT=BO+AT-\frac{1}{2}=\frac{1}{4}+1-\frac{1}{2}=\frac{3}{4}.$ Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line. Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle. Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have $${\left(r+\frac{3}{4}\right)}^2+{\left(\frac{1}{4}\right)}^2=(1-r{)}^2$$ Solving this equation gives us $$r=\boxed{\frac{3}{28}}$$