Concentration of gas above aqueous ammonia
If the container is open, as shown below, then you have a total mess. 



Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous. 

If the container is closed, then you have a somewhat more tractable problem. 



However there are two equilibriums to consider. First the gas phase must be saturated with water vapor, and second the ammonia must be distributed between the gas and liquid phases (if any liquid phase remains).  

Just out of curiosity, 3x3x1 inchs has a volume of 9 cubic inches, or about 147 mL. So if the air in the lab was at $\pu{20 ^\circ C}$ and completely dry then about $\pu{2.5 \mu L}$ of water would be needed to saturate the volume to 100% humidity. 


  Calculation:
  water vapor pressure at $\pu{20 ^\circ C} = \pu{17.5 torr}$
  atmospheric vapor pressure = 760 torr
  molar volume at $\pu{20 ^\circ C}$ and 760 torr = 24.1 liters
  Moles water = $\dfrac{17.5}{760}\cdot \dfrac{0.147}{24.1} = \pu{1.4\cdot10^{-4} moles}$
  water is $\pu{18 g/mole}$ and $\pu{1 g/ml} = \pu{0.001 g/\mu L}$
  $ \pu{\mu L water} = \dfrac{\pu{1.4\cdot10^{-4} moles}\cdot \pu{18 g/mole}}{\pu{0.001 g/\mu L}} = \pu{2.5 \mu L} $ 


NOTE: If the container is sealed there is still the problem of estimating how long it takes the system to reach equilibrium.