...
wait. I just realized there's a much easier way to do this. We're creating matter. Energy and matter are equivalent. It doesn't matter
what we're making, it only matter how much
mass we're creating.
According to
http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence 25 kilowatt-hours corresponds to 1 microgram.
Six ounces of water corresponds to
177.44 grams. That's not accounting for temperature, but it happens that the formal definition of a
gram is weight of specified volume of water at a specified temperature...so that conversion should be easy when we do it. Proceeding for casual estimate...
There are
1000000 micrograms per gram, so 1000000 * 177.44 = 177,440,000 micrograms of water in 6 ounces.
177,440,000 * 25 = 4,436,000,000 kw/hours
A
teabag is 2 grams, but the vast majority of that doesn't make it into the tea. I suspect that if we weighed 6oz of water before and after soaking a teabag in it, our instruments would probably not be sensitive enough to detect the difference. So we don't even need to worry about the mass contributed by the tea.
So now we just need to identify the mass of the teacup, convert it into kilowatt hours, add it to the above and find out how much light of our chosen wavelength that much energy is.
EDIT for fun:
Kerosene has an
energy density of 33 megajoules per liter.
http://en.wikipedia.org/wiki/Joule1 kw/h = 3.6 MJ
4,436,000,000 kw/hours = 15,969,600,000 megajoules
15,969,600,000 megajoules / 33 = 5,323,200,000
So, assuming 100% conversion efficiency, if we burned 5,323,200,000 liters of kerosene, that would provide enough energy to make a cup of tea, not including the cup.