answer is (b)
Put $x-\pi=t$ then limit $0$ changes to $-\pi$ and upper limit $2\pi$ changes to $\pi$.
$\frac{d}{dx}(x-\pi)=dt \implies dx =dt$
Integration of $t^2\sin t dt$ for limit $-\pi$ to $\pi$. One is an odd function and one is even and product of odd and even functions is odd function and integrating an odd function from the same negative value to positive value gives 0.