Expand $\sin (x-t)=\sin x\cos t-\cos x\sin t$. The integral becomes

$\sin x{\int }_0^x{t}^2\cos tdt-\cos x{\int }_0^x{t}^2\sin tdt$.

Using integration by parts (or directly differentiating the convolution form), one finds

${\int }_0^x{t}^2\sin (x-t)dt=x^2-2+2\cos x$.

(Quick check via differentiation: differentiating both sides w.r.t. $x$ twice consistently yields $\sin x$ inside the integrand convolution.)

Setting this equal to $x^2$: $-2+2\cos x=0\Rightarrow \cos x=1\Rightarrow x=2k\pi$, $k=0,1,2,\dots$.

Require $x\in [0,100]$: $2k\pi \le 100\Rightarrow k\le \frac{50}{\pi }\approx 15.91$.

So $k=0,1,2,\dots ,15$, giving 16 values.