Since there are three circles, and four sides, by the PHP, there has
to be two adjacent sides of the square $S$ that is tangent to the same circle.
WLOG, assume that is $C_1$, and the point where these two tangents meet
is $A$. Let's label $\tau_1, \tau_2, \tau_3$ be the centers of
circle $C_1, C_2, C_3$ respectively such that $\tau_2$ is in $\triangle ABC$
and $\tau_3$ is in $\triangle ACD$.
Consider two tanget points $P$ and $Q$ of $C_1$ to the square. (Specifically,
let $P$ be on side $AB$ and $Q$ on $AD$.) Because they
form a right angle, the quadrilateral $AP\tau_1Q$ is a square, and is
proportional to $\square ABCD$, therefore the points $A, \tau_1, C$ are colinear.
Moreover, this line also passes through the other point of intersection of
circle $C_2$ and $C_3$.
Now consider the feet of the perpendiculars of $\tau_2$ to side $AB$ and $BC$,
and label them $M$ and $N$ respectively. We have that $\tau_2M = \tau_2N = \tau_1P = \tau_1Q = 1$
since they are all radii of the unit circle. We also have: (1) $\tau_1\tau_2 = 1$.
(2) the triangle formed by the centers, $\triangle \tau_1\tau_2\tau_3$ is equilateral,
so $\angle \tau_3\tau_1\tau_2 = 60^\circ$. (3) Since $\angle P\tau_1Q = 270^\circ$ and
given (2) and that line $AC$ bisects it, $\angle Q\tau_1\tau_3 = \frac 12 (270^\circ - 60^\circ) = 105^\circ$.
We then have:
\begin{align}
PM &= \tau_1\tau_2 \cos(180^\circ - \angle Q\tau_1\tau_2) \\
&= \cos(180^\circ - \angle Q\tau_1\tau_3 - \angle \tau_3\tau_1\tau_2) \\
&= - \cos (\angle Q\tau_1\tau_3 + \angle \tau_3\tau_1\tau_2) \\
&= - \cos (165^\circ) = \frac{\sqrt{6}+\sqrt 2}4.
\end{align}
Finally, we have that $AB = AP + PM + MB$. We know $MB = AP = 1$,
so the length of the square is $2 + \frac{\sqrt{6}+\sqrt 2}4$.