Answer
(a) $P = 3.85\times 10^{26}~W$
(b) $I = 589~W/m^2$
Work Step by Step
(a) We can find the power output of the sun:
$I = \frac{P}{4\pi~d^2}$
$P = 4\pi~d^2~I$
$P = (4\pi)~(1.5\times 10^{11}~m)^2~(1360~W/m^2)$
$P = 3.85\times 10^{26}~W$
(b) We can find the intensity of sunlight on Mars:
$I = \frac{P}{4\pi~d^2}$
$I = \frac{3.85\times 10^{26}~W}{(4\pi)~(2.28\times 10^{11}~m)^2}$
$I = 589~W/m^2$
