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

Total power radiated by the sun is given by $P = S \times 4\pi R^2$, where $S$ is the solar constant and $R$ is the distance between the sun and the earth.

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

$P = 1400 \times 4\pi \times 2.25 \times 10^{22} = 1.26\pi \times 10^{26} \text{ W}$

The rate of mass decay of the sun is obtained using Einstein's mass-energy equivalence, $\dfrac{dm}{dt} = \dfrac{P}{c^2}$.

$\dfrac{dm}{dt} = \dfrac{1.26\pi \times 10^{26}}{(3 \times 10^8)^2} = \dfrac{1.26\pi \times 10^{26}}{9 \times 10^{16}} = 1.4\pi \times 10^9 \text{ kg s}^{-1}$

The time the sun will last is $t = \dfrac{M}{dm/dt}$, where $M$ is the present mass of the sun.

$t = \dfrac{2 \times 10^{30}}{1.4\pi \times 10^9} = \dfrac{1}{0.7\pi} \times 10^{21} \text{ s} \approx 4.55 \times 10^{20} \text{ s}$

Converting the time from seconds to years using $1 \text{ y} \approx 3.15 \times 10^7 \text{ s}$:

$t = \dfrac{4.55 \times 10^{20}}{3.15 \times 10^7} \text{ y} \approx 1.44 \times 10^{13} \text{ y}$