a) Total power output: 
b) The relative percentage change of power output is 1.67%
c) The intensity of the radiation on Mars is 
Explanation:
a)
The intensity of electromagnetic radiation is given by

where
P is the power output
A is the surface area considered
In this problem, we have
is the intensity of the solar radiation at the Earth
The area to be considered is area of a sphere of radius
(distance Earth-Sun)
Therefore

And now, using the first equation, we can find the total power output of the Sun:

b)
The energy of the solar radiation is directly proportional to its frequency, given the relationship

where E is the energy, h is the Planck's constant, f is the frequency.
Also, the power output of the Sun is directly proportional to the energy,

where t is the time.
This means that the power output is proportional to the frequency:

Here the frequency increases by 1 MHz: the original frequency was

so the relative percentage change in frequency is

And therefore, the power also increases by 1.67 %.
c)
In this second case, we have to calculate the new power output of the Sun:

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

Now we can find the radiation intensity with the equation

Where the area is

And substituting,

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