1. Let $a=\frac{\pi}{2}.$ Then the limit becomes $\lim_{x\to a}\frac{1}{(x-a)^2}\int_{x^3}^{a^3}\cos\left(t^{1/3}\right)\,dt.$

2. Define $F(x)=\int_{x^3}^{a^3}\cos\left(t^{1/3}\right)\,dt.$ We need $\lim_{x\to a}\frac{F(x)}{(x-a)^2}.$

Since at $x=a$, $F(a)=\int_{a^3}^{a^3}\cos\left(t^{1/3}\right)\,dt=0,$ we check the first derivative.

3. Differentiate $F(x)$ using the Leibniz rule: $F'(x)=-\cos\left((x^3)^{1/3}\right)\cdot \frac{d}{dx}(x^3).$ Since $x\to \frac{\pi}{2}>0$, we have $(x^3)^{1/3}=x$. Hence $F'(x)=-3x^2\cos x.$ Then $F'(a)=-3a^2\cos a= -3a^2\cdot 0=0.$

So both numerator and denominator vanish to second order.

4. Therefore, $\lim_{x\to a}\frac{F(x)}{(x-a)^2}=\frac{F''(a)}{2},$ provided $F''(a)$ exists.

Now differentiate again: $F'(x)=-3x^2\cos x,$ so $F''(x)=-6x\cos x+3x^2\sin x.$ Therefore at $x=a=\frac{\pi}{2}$, $F''(a)=-6\left(\frac{\pi}{2}\right)\cos\frac{\pi}{2}+3\left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2}.$ Using $\cos\frac{\pi}{2}=0$ and $\sin\frac{\pi}{2}=1$, $F''(a)=3\cdot \frac{\pi^2}{4}=\frac{3\pi^2}{4}.$ Hence $\lim_{x\to a}\frac{F(x)}{(x-a)^2}=\frac{1}{2}\cdot \frac{3\pi^2}{4}=\frac{3\pi^2}{8}.$

5. Therefore the correct option is $\boxed{\frac{3\pi^2}{8}}$ which is option B.