Limit. Floor. Sum.

$\lfloor x \rfloor = x - \{x\}$

$I = \displaystyle \lim_{x \to 0^{+}} \left( \lim_{n \to \infty} \dfrac{ \sum_{r = 1}^n r^2 \times x^x}{n^3} - \dfrac{ \sum_{r = 1}^n \{ r^2 \times x^x\}}{n^3}\right)$

$\displaystyle \lim_{n \to \infty} \dfrac{ \sum_{r = 1}^n r^2 \times x^x}{n^3} = \dfrac{ n(n+1)(2n + 1)}{6n^3} = \lim_{n \to \infty} \dfrac{ ( 1 + \dfrac{1}{n})(2 + \dfrac{1}{n})}{6} = \dfrac{1}{3}$

$Now~, 0 \leq \{ r^2 \times x^x\} < 1$

$\displaystyle \lim_{n \to \infty} \dfrac{ \sum_{r = 1}^n \{ r^2 \times x^x\}}{n^3} = 0$

Thus , $I = \lim_{x \to 0^{+}} \dfrac{x^x}{3}$

$\lim_{x \to 0^{+}} x^x = e^{xlogx} = e^{ \lim_{x \to 0^{+}} \dfrac{logx}{x}} = \lim_{x \to 0^{+}} \dfrac{1}{e^x} = 1$ ( as it forms $\dfrac{\infty}{\infty}$ so we can apply L.H )

$\lfloor \dfrac{1000}{3} \rfloor = \lfloor 333.333 \rfloor = 333$

use sandwich theorem

i use sandwich theorem

$\lim_{x\to0^+} x^x = e^ {\Big( lim_{x\to o^+} \dfrac{\log x}{1/x} \Big)} = 1$

(As $lim_{x\to o^+} \frac{\log x}{\frac{1}{x}} = 0$ , by applying L'Hôpital's rule )

$\Rightarrow I = \lim_{n\to \infty} \frac{\sum_{r=1}^n \lfloor{r^2}\rfloor}{n^3}$

$\Rightarrow I = \lim_{n \to \infty} \sum_{r=0}^n \Big(\frac{r}{n} \Big)^2 \frac{1}{n}$

$\Rightarrow I = \int_0^1 t^2 dt$

(Integral as limit of Riemann Sum)

$\Rightarrow I = \frac{t^3}{3} \Big| _0^1 = 1/3$

$\Rightarrow \lfloor{1000I}\rfloor = \color{#20A900}{\boxed{333}}$