Imagine that a reactor converts all the given mass into energy and that it operates at a power level of $$10^9$$ Watt. The mass of the fuel consumed per hour, in the reactor, will be: (velocity of light, $$c$$ is $$3 \times 10^8$$ m s$$^{-1}$$)

We know that electrical power is the rate at which energy is produced or consumed. Mathematically, the definition is written as

$$P \;=\; \dfrac{E}{t},$$

where $$P$$ is power in joule per second (watt), $$E$$ is energy in joule, and $$t$$ is time in seconds. In the present reactor the steady power output is given as

$$P \;=\; 10^9 \text{ W} \;=\; 10^9 \dfrac{\text{J}}{\text{s}}.$$

For one hour of operation the time interval equals

$$t \;=\; 1 \text{ h} \;=\; 60 \text{ min} \;=\; 60 \times 60 \text{ s} \;=\; 3600 \text{ s}.$$

Substituting the given power and the time of one hour into the power formula, we obtain the total energy liberated in that hour:

$$E \;=\; P \times t \;=\; \bigl( 10^9 \, \dfrac{\text{J}}{\text{s}} \bigr) \times \bigl( 3600 \, \text{s} \bigr).$$

Carrying out the multiplication step by step, we write

$$10^9 \times 3600 \;=\; 10^9 \times 3.6 \times 10^3 \;=\; 3.6 \times 10^{12}.$$

So

$$E \;=\; 3.6 \times 10^{12} \text{ J}.$$

Now, the problem states that all of the mass of the nuclear fuel is converted into energy. The famous mass-energy equivalence relation, first stated by Einstein, is

$$E \;=\; m c^2,$$

where $$m$$ is the mass that disappears and $$c$$ is the speed of light in vacuum. Rearranging this equation to solve for the mass gives

$$m \;=\; \dfrac{E}{c^2}.$$

We substitute the numerical values:

• $$E = 3.6 \times 10^{12} \text{ J},$$
• $$c = 3 \times 10^8 \text{ m s}^{-1},$$ so $$c^2 = \bigl(3 \times 10^8\bigr)^2 = 9 \times 10^{16} \text{ m}^2 \text{s}^{-2}.$$

Putting these into the rearranged formula, we get

$$m \;=\; \dfrac{3.6 \times 10^{12}}{9 \times 10^{16}} \;=\; \dfrac{3.6}{9} \times 10^{12 - 16} \;=\; 0.4 \times 10^{-4} \;=\; 4.0 \times 10^{-5} \text{ kg}.$$

Because the answer choices are expressed in grams, we convert kilograms to grams using the relation $$1 \text{ kg} = 1000 \text{ g}$$:

$$m \;=\; 4.0 \times 10^{-5} \text{ kg} \times 1000 \, \dfrac{\text{g}}{\text{kg}} \;=\; 4.0 \times 10^{-2} \text{ g}.$$

This value matches exactly the numerical amount presented in option C.

Hence, the correct answer is Option C.