You're floored! again !

$\dfrac{10^{20000}}{10^{100}+3}\\=\left(10^{19900}-3\times{10^{19800}}+9\times{10^{19700}}+\ldots+3^{198}\times{10^{100}}-3^{199}\right)+\dfrac{3^{200}}{10^{100}+3}$ .

Now, look at the last term, since $~~~3^{200}=9^{100}<10^{100}+3$ , then $\lfloor{\dfrac{3^{200}}{10^{100}+3}}\rfloor=0$ . So,

$\lfloor{\dfrac{10^{20000}}{10^{100}+3}}\rfloor=\left(10^{19900}-3\times{10^{19800}}+9\times{10^{19700}}+\ldots+3^{198}\times{10^{100}}\right)-3^{199}$ .

It's clear that the unit digit of all of numbers in the bracket is $0$ , so that the unit digit

of $\lfloor{\dfrac{10^{20000}}{10^{100}+3}}\rfloor$ is equivalent to the unit digit of $\left(-3^{199}\right)$

Now, we shall work with modulo $10$ to determine the unit digit of $\left(-3^{199}\right)$ .

$\left(-3^{199}\right)\equiv(-1)\left(9\right)^{99}\times{3}\equiv(-1)\left(-1\right)^{99}\times{3}\equiv3\pmod{10}$