Skorzystamy z twierdzenia sinusów

$2R=\frac{a}{\sin \alpha }$

 

$2R=\frac{\left|AB\right|}{\sin \sphericalangle BCA}\mid :2$  

$R=\frac{\left|AB\right|}{2\sin \sphericalangle BCA}$

$R=\frac{6\sqrt{3}}{2\sin {120}^{\circ }}$

$R=\frac{6\sqrt{3}}{2\sin \left({180}^{\circ }-{60}^{\circ }\right)}$

$R=\frac{6\sqrt{3}}{2\sin {60}^{\circ }}$

$R=\frac{6\sqrt{3}}{\cancel{2}\frac{\sqrt{3}}{\cancel{2}}}$

$R=\frac{6\sqrt{3}}{\sqrt{3}}$  

$R=6$

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Prawidłowa odpowiedź to B.