Question

At what distance from the sun is the intensity of sunlight nine times the value at the earth? (the average earth–sun separation is 1.496 1011 m.)

Answer 1

Sunlight is type of EM wave. The intensity of EM follows inverse square law.
$\frac{1}{ n^{2} }$
This means that the intensity at a distance that is n times greater will be n^2 times smaller.
Now we set this equation to be equal to 9 and we solve it for n:
$9= \frac{1}{ n^{2} } \\ \\ 9 n^{2} =1 \\ \\ n^{2} = \frac{1}{9} \\ \\ n= \frac{1}{3}$

The intensity is 9 times greater at a distance that is equal to a 1/3 of Earth's distance:
$\frac{1}{3} *1.496* 10^{11} m=4.99* 10^{10} m$