Unfortunately, option 2) is the case. The amount of fb that will be extracted, is proportional to the ratio of extractant volume (in this case xylene) to the volume of the solution containing the desired fb.
For example, if the l-ephed fb would be equally soluble in both water and xylene, extracting with twice the volume of the aqueous solution (2 parts xylene, 1 part water), would yield 66% of the fb in the xylene layer - this is the absolute maximum after sufficently mixing the layers. Doing the smae process again would get you 66% of the remaining 33% fb (thats 22% of the original fb amount), increasing the yield to 66%+22%=88%.
In reality, the extraction yield is also proportional to the ratio of the fb solubility in water to the solubility in xylene. (Anyone got the solubiliy values at hand?).
Extracting meth fb from water with a np solvent would require only a fractional volume of the original solution, since the meth fb is almost completly insoluble in water, while it is very soluble in xylene.
But since l-ephed fb (contrary to d-pseudopfed) is quite soluble in water, it would probably take several extractions with at least equal, or even larger volumes of xylene, to get the mjority of the l-ephed fb out of the water.
Which is pretty impractical, unless one got easy access to large quantities of high purity xylene, and is not concerned about handling several litres of a volatile, toxic, flammable, and environmentallly hazardous substance...