A

To compute the work done during the vaporization of water, we'll use the formula for work done during expansion against an external pressure:
Work (W) = P \Delta V
Where:
- P = external pressure (in N/m\textsuperscript{2})
- \Delta V = change in volume (in m\textsuperscript{3})

First, we need to convert the initial and final volumes into the appropriate units.
- Initial volume of water (V_i) = 1 cm\textsuperscript{3} = 1 \times 10^{-6} m\textsuperscript{3}
- Final volume of steam (V_f) = 1671 cm\textsuperscript{3} = 1671 \times 10^{-6} m\textsuperscript{3}

Now, calculating the change in volume \Delta V:
\Delta V = V_f - V_i = (1671 - 1) \times 10^{-6} = 1670 \times 10^{-6} m\textsuperscript{3}

Given:
P = 1 atm = 1.013 x 10\textsuperscript{5} N/m\textsuperscript{2}

Now substitute the values into the work formula:
W = P \Delta V = (1.013 \times 10^{5}) (1670 \times 10^{-6})
W = 1.013 \times 1670 \times 10^{-1} = 1685.71 J
To convert joules to calories (1 cal = 4.184 J):
Energy = \frac{1685.71}{4.184} = 402.32 cal
The work done during the expansion of water to steam at one atmosphere pressure is approximately 402.32 cal.