Buffering a constant current source IC to increase maximum voltage rating
C and D are the right idea.  The lower the base voltage the better, but you have to make sure the emitter voltage is within the constant current sink's compliance range.

With 1 V on the base, the emitter will only be at 300 mV or so.  If your constant current sink can still regulate current down to that voltage, then go ahead and do that.  Otherwise, set the fixed base voltage of the transistor to the minimum compliance voltage plus the B-E drop, plus a little margin.  Making the base voltage higher cuts into the voltage range the load could use, but current regulation will still be good.

Of course the transistor needs to be rated for the full voltage.

Another trick if 1 V on the base is too low but 5 V eats up too much voltage, is to put one or more of the LEDs on the emitter side of the transistor.  For example, let's say these are white LEDs that drop 3.2 V each.  With 5 V on the base, 700 mV or so for the B-E drop, and 3.2 V on the LED, that leaves 1.1 V for the current sink to work with.

Note that case B might be workable if you can be sure the LEDs always drop enough voltage for the chip to not see more than 17 V.  That will be the case when they are on and you've put enough of them in series.  The problem is when they are off.  A bleeder resistor from the bottom LED to ground might work if you can tolerate the LEDs being slightly on.  You will notice 100 &micro;A in a dark room, but in some applications that may not matter.  Two resistors making a voltage divider could cut the LED off current to almost nothing but still keep the off voltage of the bottom of the string to 17 V or less.

Added about new configuration G:

The base resistor does nothing useful.  No, it doesn't minimize base current somehow.  The base current will be what it needs to be to support the emitter current.  This is governed by the gain of the transistor.  All that resistor does is drop voltage, which makes it harder to adjust the fixed base voltage so that the emitter is just above the minimum compliance voltage of the current sink.

Also note that in configuration F (using N channel MOSFET), that the gate voltage may need to be higher.  It should be high enough so that the FET can fully turn on at maximum current.  This is a function of the G-S voltage.  The fixed gate voltage minus the minimum current sink compliance voltage therefore needs to be large enough to fully turn on the FET.