Calculate

A GPS satellite develops a fault whereby it moves into an orbit 19 800 km above the Earth’s surface and also emits a weak signal whose power is only 20 W. What will be the power per square metre of the signal received on the Earth’s surface? 

$\begin{array}{rcl} I &=& \frac{P}{4\cdot \pi\cdot r^2} \qquad P = 20\ \text{watt} \qquad r = 19\ 800\ 000\ m\\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& \frac{20}{4\cdot \pi\cdot 19800000^2} \\ I &=& 4.0596608278\cdot10^{-15}\ \frac{W}{m^2}\\ \end{array}$

If it emits the power equally in all directions then by the time it reaches the Earth it will be spread out over the surface of a sphere of radius 19800 km.  The surface area of this is 4*pi*(1.98*10^7)^2 m^2.  Divide the 20W by this area to get the power in Watts per square metre (it will be a very small number!).

Hi Many thanks for you reply , I forgot to add: 

Choose the nearest value from the list below. (W/m2 represents ‘watts per square metre’.)

Select one:

8 × 10−5 W/m2

8 × 10−8 W/m2

4 × 10−15 W/m2

2 × 10−16 W/m2

4 × 10−17 W/m2

4 × 10−9 W/m2

2 × 10−4 W/m2

sorry I forgot to add the following for the above post.

Select one:

4 × 10−15 W/m2

4 × 10−9 W/m2

8 × 10−5 W/m2

4 × 10−17 W/m2

2 × 10−16 W/m2

2 × 10−4 W/m2

8 × 10−8 W/m2