Earlier quoted context omitted.
Numbers matter. 45% of the light is in the photosynthetic active wavelength range in a normal farm vs ~20% solar panels w/ transmission conversion losses etc + 40% LED = ~8% of the incoming light is now useful for photosynthesis. So, you need a huge increase in efficiency ~400+% just to break even. PS: Not that their using solar power, but it's still worth considering.
What does the 45% of light that is in the photosynthetic range have to do with the with the efficiency of solar cells and LEDs? It would be more direct to compare % of natural light that is in photosynthetic range with % of LED light that is in that range. If you have a formula on which you're basing your assumptions then write it down so I can understand and examine your logic without any misunderstanding. Else, you…
Sunlight > photosynthisis (45%) = 45%.
Sunlight > solar cell (22%) > conversion, storage, transmission (90%)
> LED (40%) > photosynthisis(100%) = 8%
reality is LED efficecy drops over time and LED light is not quite 100% ect but this is the optomistic best case numbers.