\(284\) megagrams can be written as \(284 \times 10^6\) grams.
The SI unit of mass is kilograms, not grams. To convert grams to kilograms, divide by \(10^3\) (or multiply by \(10^{-3}\)). So, \(284 \times 10^6\) grams is equal to \(\frac{284 \times 10^6}{10^{3}}\, \text{kg}\text{.}\)
Simplifying this gives us \(284 \times 10^{3}\, \text{kg}\text{.}\)
Finally, we can express this in scientific notation as \(2.84 \times 10^{5}\, \text{kg}\text{.}\)
Therefore, the mass of fuel used by the main engine during launch is \(2.84 \times 10^{5}\, \text{kg}\text{.}\)