A 1-hectare field contains 2500 bean plants having a total surface area of 557 `m^2.` Ambient temperatures is 21 °C, winds are calm, and humidity...

Hello!

You didn't specify what "step 2 of the Analysis" is and the average loss calculated. I suppose that this loss has some numeric value. We will denote it as `I.`

Note that this loss should include time in its units, since it is the loss of water from the unit of an area over a unit of time. So the average loss is  `I` `(mL)/(m^2*h).`

Then the answer is simple. If each plant is...

Hello!

You didn't specify what "step 2 of the Analysis" is and the average loss calculated. I suppose that this loss has some numeric value. We will denote it as `I.`

Note that this loss should include time in its units, since it is the loss of water from the unit of an area over a unit of time. So the average loss is  `I` `(mL)/(m^2*h).`

Then the answer is simple. If each plant is transpiring at the average loss of `I,` then all plants are transpiring each hour:

I * (1 hour) * (the total surface area of plants) = I*557 (mL).

Probably, the result will be relatively large. We may divide it by 1000 to obtain the result in liters (L).  If the number found for `I` corresponds to the loss per day, then we should divide this formula by 24 (hours in a day).