The formula for finding the energy of a photon is;
E_gamma=(hc)/lamda
Where h is Planck's constant, c is the speed of light in a vacuum, and lamda is the wavelength of the photon. The energy of photons is typically expressed in terms of electron volts, eV, where 1 "eV" = 1.602*10^-19 "J" is the amount of energy required to accelerate an electron through one volt.
Planck's constant in terms of electron volts is 4.136xx10^-15 "eV s", and the speed of light in a vacuum is 2.998 xx 10^8 "m/s", so the energy for a single photon is;
E_gamma = ((4.136 xx 10^-15"eV s")(2.998*10^8 "m/s"))/(1.670 * 10^-6 "m")
E_gamma = .7425 "eV"
We are given the rate, nu = 9xx10^7 "/s", at which photons are received, so the rate at which energy is received can be stated as;
p = E_gamma*nu
p=(.7425 "eV")(9*10^7"/s")
p=6.682xx10^7 "eV/s"
Now we can calculate the total energy received over an hour. There are 3600 seconds in an hour.
E_T = p* t
E_T = (6.682 xx 10^7 "eV/s")(3600 "s")
2.406*10^11 eV
For comparison, this is about 1 billionth of the energy given off by a 100 watt light bulb each second.