The correct answer is $\boxed{\pi}$.
To evaluate the integral, we can use integration by parts. Let $u=x^2$ and $dv=\cos x \,dx$. Then $du=2x \,dx$ and $v=\sin x$. Substituting into the formula for integration by parts, we get
$$\int\limits_0^\pi {{{\text{x}}^2}\cos \,{\text{x dx}}} =x^2 \sin x \bigg|_0^\pi -\int\limits_0^\pi 2x \sin x \,dx$$
The first term on the right-hand side is $0$, since $x^2 \sin x$ is an odd function and the limits of integration are $0$ and $\pi$. The second term on the right-hand side can be evaluated using integration by parts again, with $u=x$ and $dv=\sin x \,dx$. This gives
$$\int\limits_0^\pi 2x \sin x \,dx =x\cos x \bigg|_0^\pi -\int\limits_0^\pi \cos x \,dx$$
The first term on the right-hand side is $0$, since $x\cos x$ is an odd function and the limits of integration are $0$ and $\pi$. The second term on the right-hand side is $-\sin x$, so the integral is equal to
$$\int\limits_0^\pi {{{\text{x}}^2}\cos \,{\text{x dx}}} =0-(-\sin x) =\pi$$
Therefore, the value of the integral is $\boxed{\pi}$.