here is my nice answer:
let $C$ be the center of enclosed circle then its radius $r$ is given by set formula by HC Rajpoot
$$r=\frac{abc}{2\sqrt{abc(a+b+c)}+ab+bc+ca}$$
where, $a=1, b=2, c=3$ are radii of three externally touching circles then
$$r=\frac{1\times2\times 3}{2\sqrt{1\times 2\times 3(1+2+3)}+1\times 2+2\times 3+3\times 1}=\frac{6}{23}$$
now, drop a perpendicular $CN$ of length $y$ from center $C$ to the x-axis to get a right triangle $C_1NC$ in which hypotenuse
$C_1C=1+\frac6{23}=\frac{29}{23}$. apply pythagorean
$$C_1N=\sqrt{(C_1C)^2-(CN)^2}=\sqrt{\left(\frac{29}{23}\right)^2-y^2}\ \ \ \ ..........(1)$$
similarly, in right triangle $CNC_2$ in which hypotenuse
$C_2C=2+\frac6{23}=\frac{52}{23}$. apply pythagorean
$$NC_2=\sqrt{(C_2C)^2-(CN)^2}=\sqrt{\left(\frac{52}{23}\right)^2-y^2}\ \ \ \ ........(2)$$
since, circle $C_1$ & $C_2$ are touching hence, $$C_1N+NC_2=C_1C_2=1+2=3$$
$$\sqrt{\left(\frac{29}{23}\right)^2-y^2}+\sqrt{\left(\frac{52}{23}\right)^2-y^2}=3$$
$$\sqrt{\frac{841}{529}-y^2}=3-\sqrt{\frac{2704}{529}-y^2}$$ taking squares on both sides, i get
$$\sqrt{\frac{2704}{529}-y^2}=\frac{1104}{529}$$
again i take square,
$$y^2=\frac{2704}{529}-\left(\frac{1104}{529}\right)^2=\frac{400}{529}$$
$$y=\frac{20}{23}$$
above is the correct value of ordinate of center $C$ of the enclosed circle
I'll work with the radii instead of the diameters. Obviously, any diameter is twice the corresponding radius. Now, note from the figure below that the triangle $OAB$ is rectangular at $A$.
Thus,
$$
\overline{OA}^{\,2} + \overline{AB}^{\,2} = \overline{OB}^{\,2}
$$
So,
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - R_1}_{\overline{AB}})^2 = (\underbrace{1 + R_1}_{\overline{OB}})^2
$$
which gives you
$$
R_1 = \frac{1}{4}
$$
Similarly, by making $B$ be the centre of the ever-smaller circles, you should be able to see that
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - 2R_1 - 2R_2 - \cdots - 2R_{n-1} - R_n}_{\overline{AB}})^2 = (\underbrace{1 + R_n}_{\overline{OB}})^2
\qquad(1)
$$
for $n > 1$. Now define
$$
S_{n} \equiv \sum_{k\,=\,1}^{n}R_k
\qquad (n \ge 1)
\qquad(2)
$$
so that $(1)$ can be written as
$$
1 + (1 - 2S_{n-1} - R_n)^2 = (1 + R_n)^2
\qquad(3)
$$
for $n > 1$. However,
$$
(1 - 2S_{n-1} - R_n)^2 =
1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n
$$
and $(3)$ gives
$$
1 + 1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n =
1 + 2R_n + R_n^2
$$
which can be simplified to
$$
(1 - 2S_{n-1})^2 = 4(1 - S_{n-1})R_n
$$
so
$$
R_n = \frac{1}{4}\frac{(1 - 2S_{n-1})^2}{(1 - S_{n-1})}
\qquad (n > 1)
\qquad (4)
$$
This gives $R_n$ in terms of the radii of only the circles with indices smaller than $n$ so it's very amenable to a recursive calculation. For example, we already have $R_1 = 1/4$. Then $S_1 = 1/4$ and $R_2 = 1/12$. Then $S_2 = 1/3$ and $R_3 = 1/24$. Then $S_3 = 3/8$ and $R_4 = 1/40$, and so on.
Computing a few more values in succession, it's easy to discern the pattern:
$$
S_n = \frac{n}{2(n+1)}
\qquad\mbox{and}\qquad
R_n = \frac{1}{2n(n+1)}
\qquad (n \ge 1)
$$
That these satisfy $(2)$ and $(4)$ can be easily proven by induction on $n$.
Best Answer
Note
$$OA = \sqrt2 r_1 = r_1 + r_2 + \sqrt2 r_2$$
leading to the ratio $\frac{r_2}{r_1} = (\sqrt2-1)^2 =a $. Likewise $\frac{r_3}{r_2}=\frac{r_4}{r_3}=…=a$. Thus, the total areas is of a geometric sum
\begin{align} S &= \pi(r_1^2 + r_2^2 + r_3^2 + … ) = \pi r_1^2(1+ a^2+ a^4 + a^6 + … ) \\ & = \frac{\pi r_1^2}{1-a^2}=\frac{100\pi}{1-(\sqrt2-1)^4} = \frac{25\pi}{3\sqrt2-4} \end{align}