Answer:
A,
Step-by-step explanation:
To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which

Where,
h = Plank's constant

c = Speed of light

=Wavelength
If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.
The larger the amplitude of the wave, the smaller the energy.
On the other hand we have the frequency value defined as

In this case the frequency is also inversely proportional to the wavelength.
In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.
In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.