Calculate water hardness from grams of CaCO3 We have to figure out what the water hardness in mg/L or ppm is for a $\pu{20ml}$ solution of $\pu{0.400M}$ $\ce{CaCl2}$. 

We learned that the formula for hardness is mg/L of calcium carbonate per liter. I started by calculating that there would be $\pu{0.801 g}$ of $\ce{CaCO3}$ precipitate if reacted with $\ce{Na2CO3}$.  I then converted $\pu{0.801 g}$ to mg and got $\pu{801 mg}$. I finally divides this by the original $\pu{20ml}$ or $\pu{0.02 L}$ but this gives $\pu{40050 mg/L}$ but the correct answer is $\pu{400 mg/L}$. 

I feel like I might be making a mistake with units as my answer is 100 times greater than the correct answer but I don't see the problem.
Unless I'm missing something, I don't see the problem with your working either. Perhaps check whether you got the original concentration of $\ce{CaCl2}$ given in the problem right (the volume of the solution does not affect your final answer). The given answer may well be wrong, it's irritating but it does happen.

There is one thing I want to comment on though: never round off your answer too early. Since you obtained the intermediate answer of $801 \text{ mg}$, I am assuming you used a precise molar mass for $\ce{CaCO3}$ - I will take this to be $100.09 \text{ g mol}^{-1}$. (If you used the value $100$ you would have obtained $800 \text{ mg}$.) Your calculations should be:


Number of moles of $\ce{CaCO3}$ formed upon addition of excess $\ce{CO3^2-}$


$$\eta_{\ce{CaCO3}} = (0.400 \text{ mol dm}^{-3})(0.02 \text{ dm}^{3}) = 0.008 \text{ mol}$$


Mass of said $\ce{CaCO3}$


$$m_{\ce{CaCO3}} = (0.008 \text{ mol})(100.09 \text{ g mol}^{-1}) = 0.80072 \text{ g} = 800.72 \text{ mg}$$


Hardness of water in $\text{mg L}^{-1}$


$$[\ce{CaCO3}] = \frac{800.72 \text{ mg}}{0.02 \text{ L}} = 40036 \text{ mg L}^{-1}$$

This is your final answer and if you want to round to the appropriate number of significant figures, you should do so only at this point. Your problem is that you rounded too early (after step 2) and this means you obtained an imprecise final answer, $40050 \text{ mg L}^{-1}$.

Here, the appropriate number of sf is technically 3, since the original concentration of $\ce{CaCl2}$ is given to you in 3 sf. Since you obtained the imprecise answer $40050 \text{ mg L}^{-1}$, you would have rounded this to $40100$. But if you stuck to the precise value at step 2 and obtained the answer $40036 \text{ mg L}^{-1}$, you would round this to $40000$.