This is just a revision of this calc done by Drite77 a few months ago. Credits to them, first and foremost, since all I'm doing is plugging a different value into more or less the same process. For easy reference, the feat can be seen here.
According to the evaluation, this site no longer uses the horizon distance calculator that Drite used, so I'm going to be plugging in the average horizon distance we now use, which means 20000 meters for the diameter for the cloud feat according to DemonGodMitchAubin's comment.
Edit: This calc has been edited as per Ugarik's comment, which corrected me in that 20000 meters is the radius, as opposed to the diameter.
Edit 2: This calc has been edited again as per pointed out in a CRT of me having forgotten to convert grams to kilograms. Whoops.
I'll be making the same assumptions as Drite, so we'll assume the clouds created are Nimbostratus, which average at 3km in thickness.
Using a cylinder for cloud volume:
Volume = π*(r^2)*h, where r is radius and h is height (in this case, the thickness of the clouds).
Volume = π*(20000^2)3000
Volume = 3,769,911,184,307.7518861551720599354 m^3
We'll be using condensation, since Razor created the clouds; which means we'll be multiplying the mass of the water in the clouds by the latent heat of vaporisation of water, which is 2264705 J/kg. Water density in Nimbostratus clouds is 1 g/m^3, so with that:
Condensation Value = Water Mass x Latent Heat of Vaporisation of Water
Condensation Value = 3,769,911,184.3077518861551720599354 kg x 2264705 J/kg
= 8,537,736,708,657,687.235335048939996 Joules
= 8.537736708657687235335048939996e15 Joules
= ~8.538e15 Joules
OR
= 2.0405680470023153106 Megatons of TNT
This gives us an approximate value of 8.538e15 Joules or 2.04 Megatons of TNT, Low 7-B.