Selecting an inductor based on schematic
40 mW divided by 3.3 V implies a current of about 13 mA. So choose any size rated for at least twice this amount for margin with low resistance and the cheapest price.

https://www.digikey.com/product-detail/en/wurth-electronics-inc/744032680/732-3060-6-ND/2626313

When an inductor exceeds it rated current, it will lose its inductance, like a spring bottoming out so it acts as a noise filter or low pass filter if there is any noise on the  3.3 V supply. Low supply noise is important for the liquid paper display.

Rev A

It is not clear to me why or where an iron core inductor was specified but these tend to have higher permeability and thus less wire and lower resistance. The reason I mention this is when switching the inductor the current rise   is due to the ratio of L/R=T time constant. The max current is Vcc/ESR of the total resistance but the current must be only pulsed for a low duty cycle such that the average current is only . The FET is &lt; 0.3 Ohms the series source is 3 Ohms so I suspect the inductor needs to be similar to the R values of iron core parts with 68uH which is &lt; 0.5 Ohms but it does not have to be iron core, per se. 


  They should have specified part numbers in the BOM.


Here is my Rev A choice https://www.digikey.com/products/en/inductors-coils-chokes/fixed-inductors/71

I have changed my selection based on &lt;=0.5 Ohm rather than >> 13mA so you will find it rated for much higher currents, even tho it will be limited by the total equivalent series resistance (ESR) of R(L1+Q1+R2)= 0.5+0.3+3 = 3.8 Ohms thus current ramp is 68uH/3.8 Ohms = 17 us. (??) and if pulsed for this duration the ramp current rises towards 3.3V/3.8 Ohms = 870 mA To use less than 13 mA average ( assuming that is correct) the pulse duty cycle must be 13/870*100=1.5%  hmm