1. **State the problem:** Given the mass of one water molecule $m(\mathrm{H_2O}) = 3.01 \times 10^{-26}$ kg, the amount of substance $n = 6.6 \times 10^{-3}$ mol, and Avogadro's number $N_A = 6.02 \times 10^{23}$ mol$^{-1}$, find the total mass of the water sample. 2. **Formula used:** The total mass $m_{total}$ of a sample can be found by multiplying the number of molecules by the mass of one molecule: $$m_{total} = N \times m(\mathrm{H_2O})$$ where $N$ is the total number of molecules. 3. **Calculate the total number of molecules:** $$N = n \times N_A = 6.6 \times 10^{-3} \times 6.02 \times 10^{23}$$ 4. **Multiply the numbers:** $$N = (6.6 \times 6.02) \times 10^{-3 + 23} = 39.732 \times 10^{20} = 3.9732 \times 10^{21}$$ 5. **Calculate total mass:** $$m_{total} = 3.9732 \times 10^{21} \times 3.01 \times 10^{-26}$$ 6. **Multiply the coefficients and add exponents:** $$m_{total} = (3.9732 \times 3.01) \times 10^{21 - 26} = 11.95 \times 10^{-5} = 1.195 \times 10^{-4} \text{ kg}$$ **Final answer:** $$\boxed{m_{total} = 1.195 \times 10^{-4} \text{ kg}}$$