Because

$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^{n}\left(k\sin\frac{k\pi}{n}\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\left(\frac{k}{n}\sin\frac{k}{n}\pi\right)\right)$$

therefore the desired result is the Riemann integral of function $f(x)=x\sin(x\pi)$ over $[0, 1]$, i.e.

$$\begin{align*} \int_0^1 x\sin(x\pi)dx&=\frac{1}{\pi}\int_0^1 x\sin(x\pi)d(x\pi)\\ &=-\frac{1}{\pi}\int_0^1 xd\cos(x\pi)\\ &=-\frac{1}{\pi}\left(\left. x\cos(x\pi)\right|_0^1-\int_0^1\cos(x\pi)dx\right)\\&=-\frac{1}{\pi}\left(-1-\left.\frac{1}{\pi}\sin(x\pi)\right|_0^1\right)\\&=\boxed{\frac{1}{\pi}} \end{align*}$$