Let's represent the problem as follows, for a parabola $y^2=4ax$:

Therefore, if $OC = x_1 \rightarrow y = \sqrt {4 a x_1} = \pm 2\sqrt{a x_1}$

Therefore $A = (x_1, 2\sqrt{ax_1})$ and $B = (x_1, -2\sqrt{ax_1})$

$\Rightarrow AB = AC + BC = 2 AC = 2 \times 2 \sqrt {ax_1} = 4 \sqrt{ax_1}$

Since $\Delta\;OAB$ is equilateral $OA^2 = OB^2 \rightarrow x_1^{2} + (2\sqrt{ax_1})^2 = (4 \sqrt{ax_1})^2$

$\Rightarrow x_1^2 + 4ax_1 = 16ax_1 \rightarrow x_1^2 = 12ax_1 \rightarrow x_1 = 12a$

Therefore $AB = 4 \sqrt {a x_1} = 4 \times \sqrt {12 \times a^2} = 8 \sqrt{3}\;a$