If $M, N$ are the centres of two of the circles, and $P, Q$ are their projections onto the side $AB$ of the triangle as in the diagram below, the right angled triangles $APM$ and $BQN$ have internal angles of $30^\circ$, $60^\circ$ and $90^\circ$. Hence $|AM| = 2|MP| = 2$ since $MP$ is a radius of the circle.

From Pythagoras we get $|BQ| = |AP| = \sqrt{3}$. Since $|PQ| = |MN| = 2$ we now obtain $|AB| = a = 2 + 2\sqrt{3}$.

The area of an equilateral triangle with side length $a$ is equal to
$$\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}(2+2\sqrt{3})^2}{4} = 2(3+2\sqrt{3}).$$

Three circles of radius $1$ have a total area of $3\pi$. The fraction of the area of the triangle covered by the circles is
$$\frac{3\pi}{2(3+2\sqrt{3})} = \frac{(2\sqrt{3}-3)\pi}{2}.$$