You're floored !

$\begin{aligned} \dfrac {10^{93}}{10^{31}+3} & = \dfrac {(10^{31}+3)^3-3(3)10^{62}-3(3^2)10^{31}-3^3}{10^{31}+3} \\ & = \dfrac {(10^{31}+3)^3-9(10^{31})(10^{31}+3)-27}{10^{31}+3} \\ & = (10^{31}+3)^2 - 9(10^{31}) - \dfrac {27}{10^{31}+3} \\ & = 10^{62}+6(10^{31}) + 9 - 9(10^{31}) - \dfrac {27}{10^{31}+3} \\ & = 10^{62}-3(10^{31}) + 9 - \dfrac {27}{10^{31}+3} \\ & = 10^{62}-3(10^{31}) + 8 + \dfrac {10^{31} - 24}{10^{31}+3} \end{aligned}$

$\begin{aligned} \Rightarrow \left \lfloor \dfrac {10^{93}}{10^{31}+3} \right \rfloor & = 10^{62}-3(10^{31}) + 8 = (10^{31} - 3)10^{31} + 8 \\ & = 99999...99700...00008 \end{aligned}$

$\Rightarrow$ The sum of the units and tens digits $= 0+8= \boxed{8}$

$\begin{array}{c}&\dfrac{{10}^{93}}{{10}^{31}+3}&={10}^{62}-3\times{{10}^{31}}+9-\dfrac{27}{{10}^{31}+3}\\&={10}^{62}-3\times{{10}^{31}}+8+\dfrac{{10}^{31}-24}{{10}^{31}+3}\end{array}$

Since $~~\lfloor\dfrac{{10}^{31}-24}{{10}^{31}+3}\rfloor=0$ , then $~~\lfloor\dfrac{{10}^{93}}{{10}^{31}+3}\rfloor={10}^{62}-3\times{{10}^{31}}+8$ .

Both ${10}^{62}~$ and $\left(-3\times{{10}^{31}}\right)$ has ten and unit digit $0$ , so the unit digit of

$\lfloor\dfrac{{10}^{93}}{{10}^{31}+3}\rfloor=8$

Hence, the answer is $0+8=8$