A

Step 1: Understanding the reaction
The D-T fusion reaction can be represented as:
$$^2H + ^3H \rightarrow ^4He + n + 17.6 \text{ MeV}$$
Here, deuterium (D) and tritium (T) fuse to form helium (He) and a neutron (n), releasing a substantial amount of energy in the form of 17.6 MeV per reaction.

Step 2: Convert MeV to Joules
The energy released per fusion reaction in Joules, given that 1 MeV = 1.6 \times 10^{-13} J, is:
$$E = 17.6 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 2.816 \times 10^{-12} \text{ J}$$

Step 3: Calculate the number of reactions needed for 1 MW
The power we need is 1 MW or 1 million watts, which is equivalent to 1 million Joules per second:
$$P = 1 \text{ MW} = 10^6 \text{ J/s}$$
To find the number of reactions per second:
$$N_{reactions} = \frac{P}{E} = \frac{10^6 \text{ J/s}}{2.816 \times 10^{-12} \text{ J}} \approx 3.55 \times 10^{17} \text{ reactions/s}$$

Step 4: Calculate the amount of deuterium and tritium consumed
Each reaction consumes 1 deuterium and 1 tritium atom, so the rate at which deuterium and tritium are consumed will be the same as the number of reactions:
$$\text{Rate of D consumption} = \text{Rate of T consumption} = N_{reactions} \approx 3.55 \times 10^{17} \text{ atoms/s}$$

Since we are asked for the consumption rate in terms of mass or similar, if you need the total mass of each consumed per second, that can be calculated as:
For deuterium, the molar mass is approximately 2 g/mol:\
$$\text{Mass of D consumed/s} = \frac{N_{reactions}}{N_A} \times \text{Molar Mass of D}$$
where $N_A$ (Avogadro's number) is approximately 6.022 \times 10^{23} mol^{-1}.
Therefore, the mass of D consumed per second:
$$\approx \frac{3.55 \times 10^{17}}{6.022 \times 10^{23}} \times 2 \text{ g/mol} \approx 1.18 \times 10^{-6} \text{ g/s}$$
Similarly for tritium. However, since the question doesn't specify units, we are primarily concerned with the rate at which the fusion process occurs, which aligns with the number of atoms being consumed per second.

Conclusion
Therefore, the rate at which deuterium and tritium are consumed to produce 1 MW of energy through D-T fusion is approximately 3.55 \times 10^{17} atoms/s for each isotope.