Let $r$ be the radius of the smallest circle. Let $O_1$ be the center of the circle of radius 1, $O_2$ be the center of the circle of radius 2, $O_3$ be the center of the circle with radius 3, and $O_r$ be the center of the circle of radius $r$.
If you draw $\triangle O_1O_2O_r$, you can easily verify that it has side lengths 3, $1+r$, and $2+r$. In addition, you can also verify that the cevian $O_rO_3$ to the side with length 3 has length $3-r$. We now have enough information to solve for $r$. If we let $\theta=\angle O_1O_3O_r$, then by the Law of Cosines:
\begin{align}
4+(3-r)^2+4(3-r)\cos\theta &= (1+r)^2\\
1+(3-r)^2-2(3-r)\cos\theta &= (2+r)^2.
\end{align}
Adding the first equation to twice the second yields the equation $6+3(3-r)^2=(1+r)^2+2(2+r)^2$, which has the solution $r=\frac 67$.
Alternatively, you can use Descartes' Theorem, in particular, the following formula:
$$\frac1r=\frac 11+\frac12-\frac13\pm2\sqrt{\frac1{1\cdot2}-\frac1{2\cdot3}-\frac1{3\cdot1}}=\frac 76,$$
so $r=\frac67$.
Using coordinates . . .
For convenience of notation, let $h=\sqrt{3}$.
Let $B = (-1,0),\;C=(1,0),\;A=(0,h)$.
Let $P$ be the center of the required circle.
Then $P=(0,1+r)$, where $r$ is the unknown radius.
Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.
Let $M$ be the midpoint of segment $CA$.
Then $M=\bigl({\large{\frac{1}{2}}},{\large{\frac{h}{2}}}\bigr)$.
Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.
Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.
Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.
Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
\begin{align*}
&MP^2=(1-r)^2\\[4pt]
\implies\;&\left({\small{\frac{1}{2}}}-0\right)^{\!2}+\left({\small{\frac{h}{2}}}-(1+r)\right)^{\!\!2}=(1-r)^2\\[4pt]
\implies\;&r=\frac{4h-1-h^2}{4(4-h)}\\[4pt]
&\phantom{r}=\frac{4\sqrt{3}-4}{4(4-\sqrt{3})}\\[4pt]
&\phantom{r}=\frac{3\sqrt{3}-1}{13}\approx .3227809558\\[4pt]
\end{align*}
Best Answer
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $\triangle DGE$ and $\triangle ECF$ respectively
\begin{align*} GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rr\end{align*} Hence
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.