Explanation

(b) From Einstein’s mass energy relation the energy released is
$\mathrm{\Delta}E\ = \ \mathrm{\Delta}mc^{2}$
where$\ \mathrm{\Delta}m$ is mass and c is speed is light.
Given
$\mathrm{\Delta}m\ = \ 1mg\ = \ 1\ \times 10^{- 6}kg,\ c = 3\ \times \ 10^{8}\ m/s$
$\therefore\ \ \ \ \ \ \mathrm{\Delta}E\ = \ 1\ \times \ 10^{- 6}\ \times \ \left( 3\ \times \ 10^{8} \right)^{2}$
$\mathrm{\Delta}E\ = \ 9 \times 10^{10}\ J$
The rate at which energy is dissipated is known as power, $ie,$
$\ P = \frac{\mathrm{\Delta}\ E}{t}\ \ = \frac{9\ \times 10^{10}}{1} = 9 \times 10^{10}W$
Since, $\ \ 10^{3}\ W\ = \ 1k\ W$
$\therefore\ \ \ \ \ \ \ \ \ \ \ P = \ 9\ \times \ 10^{7}\ kW$