Correct Answer: $2\frac{2}{3}$

Solution : Given that $x\cos^{2}30^{\circ}\cdot \sin60^{\circ}=\frac{\tan^{2}45^{\circ}\cdot \sec60^{\circ}}{\operatorname{cosec}60^{\circ}}$
We know that,
$\cos30^{\circ} = \frac{\sqrt{3}}{2}$, so $\cos^{2}30^{\circ} = \frac{3}{4}$
$\sin60^{\circ} = \frac{\sqrt{3}}{2}$
$\tan45^{\circ} = 1$, so $\tan^{2}45^{\circ} = 1$
$\sec60^{\circ}= 2$
$\operatorname{cosec}60^{\circ} =\frac{2}{\sqrt{3}}$
Putting these values in the equation, we get:
$x×\frac{3}{4}×\frac{\sqrt{3}}{2}=\frac{1×2}{\frac{2}{\sqrt{3}}}$
⇒ $x×\frac{3}{4}×\frac{\sqrt{3}}{2}=\frac{2×\sqrt{3}}{{2}}$
Solving for $x$, we get:
⇒ $x=\frac{8}{3}=2\frac{2}{3}$
Hence, the correct answer is $2\frac{2}{3}$.