Hint. Consider two of those circles, and the rectangle given by 4 points: two centers of these circles, and two their tangency points on one side of the triangle. What can you tell about the proportion of the sides of the rectangle? Do you know where the two centers of the circles are placed? Does it help?
If you get something like $x+2\cdot(x\cdot\tan\frac{\pi}{6})+x=a$ then you are thinking in the right direction, I guess.
Here's a (not-to-scale) picture of the situation:
Necessarily, each circle center ($D$, $E$, or $F$) is the point where an angle bisector meets an opposite edge; moreover, the points of tangency of a circle with the adjacent edges (for instance, $D^\prime$ and $D^{\prime\prime}$) are simply the feet of perpendiculars from the center to those edges.
We'll write $a$, $b$, $c$ for the lengths of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $d$, $e$, $f$ for the radii of $\bigcirc D$, $\bigcirc E$, $\bigcirc F$. Now, observing that each angle bisector cuts the triangle with into sub-triangles with convenient "bases" and "heights", we can compute the area, $T$, of $\triangle ABC$ in three ways:
$$T \;=\; \frac12 d\;(b+c) \;=\; \frac12 e\;(c+a) \;=\; \frac12 f\;(a+b) \tag{1}$$
Of course, writing $r$ for the inradius of $\triangle ABC$, we have a well-known fourth formula for area:
$$T \;=\; \frac12 r\;(a+b+c) \tag{2}$$
We can easily eliminate $a$, $b$, $c$ from the above. For instance,
$$\begin{align}
b+c = \frac{2T}{d}\quad c+a=\frac{2T}{e}\quad a+b = \frac{2T}{f} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\;\right) \tag{3} \\
a+b+c = \frac{2T}{r} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{2}{r}\;\right) \tag{4}
\end{align}$$
so that, as @Jack notes,
$$\frac{2}{r} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \tag{$\star$}$$
Addendum (four years later!). As @jmerry has observed, the specific configuration in the problem statement is invalid. If we solve $(1)$ for $a$, $b$, $c$, we find
$$a = \left(-\frac1d + \frac1e + \frac1f \right)T \qquad b = \left(\frac1d - \frac1e + \frac1f \right)T \qquad c = \left(\frac1d + \frac1e - \frac1f \right)T$$
With $d=18$, $e=6$, $f=9$, these become $a=2T/9$, $b=0$, $c=T/9$, which do not correspond to a valid triangle ... not even a validly-degenerate one. (It's a good thing I didn't claim my picture was to scale.) A valid triangle requires that the three aspects of the Triangle Inequality hold
$$a \leq b+c \qquad b \leq c+a \qquad c \leq a+b$$
which, in turn, require
$$\frac3d \geq \frac1e + \frac1f \qquad \frac3e \geq \frac1f+\frac1d \qquad \frac3f \geq \frac1d+\frac1e$$
(The first of these is violated by the given values of $d$, $e$, $f$.)
Best Answer
Let the unknown radius be $r$. Then the vertical distance between the centres of $B,C$ is $r-9$, hence by Pythagoras their horizontal distance is $\sqrt{(9+r)^2-(r-9)^2}=6\sqrt{r}$. Likewise, the vertical distance between centres of $A,C$ is $r-4$ and therefore the horizontal distance is $\sqrt{(4+r)^2-(r-4)^2}=4\sqrt r$. Thus the horizontal distance between centres of $A,B$ is $6\sqrt r-4\sqrt r=2\sqrt r$ and the vertical distance is $\sqrt{13^2-4r}$, but it is also $2r-13$. We conclude $$13^2-4r=(2r-13)^2 $$ and so ($r=0$ or) $$r=12. $$