Solution:

The centers of the three unit circles form an equilateral triangle with side $2.$ Its center is the center of the large circle.

The distance from the center of an equilateral triangle to a vertex is $\dfrac{\text{side}}{\sqrt3} = \dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}.$

Adding the unit radius, the large radius is $1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}.$

Thus, the correct answer is D.