Just outside the earth's atmosphere, energy from the sun arrives at...

The intensity of solar radiation at the earth's distance is $I = 1400 \text{ W m}^{-2}$.

The total power radiated by the sun is distributed over a spherical surface of radius $R = 1.5 \times 10^{11} \text{ m}$.

Total power $P = I \times 4\pi R^2$

Substituting the given values:

$P = 1400 \times 4\pi \times (1.5 \times 10^{11})^2$

$P = 1400 \times 4\pi \times 2.25 \times 10^{22} = 12600\pi \times 10^{22} \text{ J s}^{-1}$

According to the mass-energy equivalence principle, $E = mc^2$, the rate of mass loss is:

$\dfrac{dm}{dt} = \dfrac{P}{c^2}$

Using $c = 3 \times 10^8 \text{ m s}^{-1}$:

$\dfrac{dm}{dt} = \dfrac{12600\pi \times 10^{22}}{(3 \times 10^8)^2}$

$\dfrac{dm}{dt} = \dfrac{12600\pi \times 10^{22}}{9 \times 10^{16}} = 1400\pi \times 10^6 \text{ kg s}^{-1}$

Taking $\pi = \dfrac{22}{7}$:

$\dfrac{dm}{dt} = 1400 \times \dfrac{22}{7} \times 10^6 = 200 \times 22 \times 10^6 = 4400 \times 10^6 \text{ kg s}^{-1}$

$\dfrac{dm}{dt} = 4.4 \times 10^9 \text{ kg s}^{-1}$