Solution:

The answer is (D). Let us call $\omega_{1}, \omega_{2}, \omega_{3}$ the first three circles and $\Omega$ the fourth one, and let $P_{1}, P_{2}, P_{3}, P$ be their respective centers. The triangle $P_{1} P_{2} P_{3}$ is evidently equilateral (each of its sides equals twice the radius of the circles $\omega_{1}, \omega_{2}, \omega_{3}$, that is, it is equal to 2). The point $P$ is equidistant from $P_{1}, P_{2}, P_{3}$, and is therefore the circumcenter of $P_{1} P_{2} P_{3}$. The distance $P P_{1}$ is then equal to the radius of the circle circumscribed about an equilateral triangle with side 2. Since in an equilateral triangle the circumcenter coincides with the centroid, and the centroid divides each median in ratio $1:2$, we have that $P P_{1}$ is also equal to $2/3$ of the median (or equivalently, the altitude) issuing from $P_{1}$. Since the angle at $P_{2}$ is $60^{\circ}$, the length of this altitude can be calculated as $\frac{\sqrt{3}}{2} P_{1} P_{2}=\sqrt{3}$. We therefore obtain that the length $P P_{1}$ is $\frac{2}{3} \sqrt{3}$. This length, however, is also equal to the sum of the radii of $\omega_{1}$ and $\Omega$, so by subtraction we obtain that the radius of $\Omega$ is $\frac{2}{3} \sqrt{3}-1$.