If the radius of each of four outer circles is $r$, then the radius of the innermost circle is

Looking at the figure, four circles of radius $r$ are arranged symmetrically, each touching its two adjacent circles. A small circle sits at the center, touching all four outer circles.

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Since adjacent outer circles touch each other, the distance between their centers is $r + r = 2r$. The four centers form a square with side $2r$.

Placing the center of the arrangement at the origin, each outer circle's center is at a distance of $r$ along both the $x$ and $y$ directions, i.e., at $(\pm r, \pm r)$.

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The distance from the origin to any outer circle's center, say $(r, r)$:

$d = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$

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The innermost circle is centered at the origin and is externally tangent to each outer circle. So the distance between their centers equals the sum of their radii:

$r\sqrt{2} = R_{\text{inner}} + r$

$R_{\text{inner}} = r\sqrt{2} - r$

$R_{\text{inner}} = r(\sqrt{2} - 1)$

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The radius of the innermost circle is $r(\sqrt{2} - 1)$.