Solution:

Since $AB = BC = AC$ by rotational symmetry, we know it is an equilateral triangle.

Then, one-fourth of $AB$ can be found as a the leg of a right triangle with hypotenuse $1$ and is opposite to the $60^{\circ}$ angle, making it $$\sin(60^\circ) = \dfrac{\sqrt 3}2.$$

As such, $$AB = 4\cdot \dfrac{\sqrt 3}2 = 2\sqrt 3.$$

Then, since it is an equilateral triangle, it has area $$\dfrac{s^2 \sqrt 3}4 = \dfrac{12\sqrt 3}4 = 3 \sqrt 3.$$

Thus, the correct answer is B .