You are on the right track but it is unclear from what you say what you are actually doing. The BU paper says terminate at 3% of C/1 rate - which is far too low for sensible use. You say you are trying to copy their approach BUT also that you are getting 20 mA etc.
You need to say WHEN you are getting 20 mA - when cells are discharged or fully charged or ...?
It helps to give battery specs AND a link.
Panasonic NCR18650B datasheet here.
If the cells are genuine Panasonic then they are nominal 3250 mAh capacity
and are specified to be charged at C/2 max = 1625 mA.
If the cells are less than fully charged (say Vpack < 6 x 4 = 24V) then applying a source able to charge at up to 1625 mA at 6 x 4.2 = 25.2V WILL produce a charge current of 1625 mA except in one case. If any cell has been discharged to a dangerously low level the internal protection circuit may limit current to a trickle up rate - often Cmax/10 = 160 mA here. This will jhopefully not be the case here.
Your cited power supply has the makings of what you need but needs to be set up correctly. If the Iamx pot is set to minimum it may limit current to the 20 mA you are seeing. So ...
Note that some ammeters have quite high resistance on mA ranges (I have seen a 17 Ohm milliameter!). 10A range is usually low Ohms.
A good method is to add a series eg 0.1 Ohm or 0.01 Ohm series resistor and measure the voltage across it.
A 0.1 Ohm will drop V = IR = 1 x 0.1 = 0.1V at 1 Amp and a 0.01 Ohm resistor will drop 0.01 V at 1 A.
As Diego says, stoppingh charge when Vbattery first rises to 4.2V/cell will only provide about 80% of full capacity BUT the cycle life may be doubled if you always stop there and 80% capacity may be fine.
ON NO ACCOUNT should you leave the voltage source connected to the battery indefinitely. A LiIon battery will have its cycle life badly reduced by not removing Vchg when Ichg has dropped to a small fraction of initial value. Imax/10 is usually fairly aggressive. Imax/4 or Imax/2 provide most of max capacity and are much kinder to the battery.