Solution
$$\sin120^{\circ\ }\cos60^{\circ\ }\cot30^{\circ\ }\operatorname{cosec}^230^{\circ\ }=\sin\left(90+30^{\circ\ }\right)\cos60^{\circ\ }\cot30^{\circ\ }\operatorname{cosec}^230^{\circ\ }$$
$$=\cos30^{\circ\ }\ \cos60^{\circ\ }\cot30^{\circ\ }\operatorname{cosec}^230^{\circ\ }$$
$$=\frac{\sqrt{\ 3}}{2}\times\ \frac{1}{2}\times\ \sqrt{\ 3}\times\ 4$$
$$=3$$
Hence, the correct answer is Option B
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