How do you correctly round this problem $\dfrac{30.00\,g}{10.0\,mL}$ ?

Hint: The density of any substance is the measure of mass upon volume. Here the given entity that has to be rounded up is density. The rounding up means simplifying the given mathematical operation or equation.

Complete answer: The problem is rounded up by dividing the given entity of mass by the given volume. For this the volume needs to be converted into liters.
We know that, 1000 mL = 1 L
So, 10.0 mL = $\dfrac{1\,L\times 10\,mL}{1000\,mL}$
Therefore, 10.0 mL = 0.01 L
Now, 10.0 mL will be written as 0.01 L in the given problem as$\dfrac{30.00\,g}{0.01\,L}$, so this problem will be solved as,
$\dfrac{30.00\,g}{0.01\,L}$= 3000 g/L or $\dfrac{30.00\,g}{10.0\,mL}$= 3.00 g/ mL
Hence, the problem is rounded up as 3000 gram per liter or 3.00 gram per milliliter.

Additional information: the problems that need to be rounded up need the concept of significant figures. Significant figures are a count of the valid figures in a numerical value. The rule of significant figures suggests that zeros after the decimal are said to be significant. Also the zero between a decimal and a non-zero digit is significant, while the zeros after or before the non-zero digit are insignificant.

Note: The given problem contains the significant figure of 4 in 30.00 g and that of 1 in 0.01 L. So, if we take out the value through significant figures, the answer will still come as 3000 g/L as all the digits in the mass of 30.00 are significant.