How would you do it?

We have that, $10(x-1)\leq \sqrt{n}<10x,\forall\ x\in Z^{+}\\ \Longrightarrow (x-1)\leq \dfrac{\sqrt{n}}{10}<x\\ \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=x-1\\ x=1 \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=0 [0 \leq n<100]\\ x=2 \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=1 [100 \leq n <400]\\ x=3 \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=2 [400 \leq n <900]\\ x=4 \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=3 [900 \leq n <1600]\\ x=5 \Longrightarrow \left[ \dfrac{\sqrt{n}}{10}\right]=4 [1600 \leq n <2016]$ ,the final answer can be calculated as follows, $100 \times 0+300\times 1+500\times 2+ 700\times 3+416\times 4=5064$ .And done!