We know that the length of the chord in a circle is $\(4 \sqrt{3}=\mathrm{AB}\)$ (as shown in the figure above).
As perpendicular drawn from the centre of the circle to the chord bisects the chord, we get $\(\mathrm{AC}=\)$ $\(C B=\frac{1}{2}(4 \sqrt{3})=2 \sqrt{3}\)$

Now, let us assume the measure of angle $\(O B C=\theta^{\circ}\)$, so in triangle $\(B O C\)$, we get $\(\operatorname{Cos} \theta=\frac{Base}{ Hypotenuse} =\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}\)$

Finally in triangle BOC , we get angle $\(\mathrm{COB}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)=180^{\circ}-120^{\circ}=60^{\circ}(\mathrm{Sum}\)$ of the angles of a triangle is 180 degrees)

As $\(\triangle \mathrm{AOC} \cong \triangle B O C\)$, we get $\(\angle \mathrm{AOC}=\angle \mathrm{BOC}=60^{\circ} \Rightarrow \angle \mathrm{AOB}=120^{\circ}=\)$ angle subtended by the chord at the centre of the circle.

Hence the answer is (E).