I don't know what's wrong with using a white card to find the focus. The focus is where you see the image of the lamp filament. You cannot focus the filament to a smaller image size than the actual filament itself without running up against the laws of thermodynamics.
EDIT: OK, this answer is really wrong (see comments below). Leaving it up because I don't believe in covering my tracks.
EDIT 2: I'm going to try this one more time. What is probably true is that no system of lenses is going to cram more light into the fibre than you would get just by butting up the end of the fibre directly against the lamp. Any takers?
EDIT 3: Second answer also wrong. Let's try this: Express the capture area of the lens as a fraction of the total emission angle of the source. That is the same fraction which gives you the minimum (optimum) image size as compared to the size of the source.
(Also, re-reading the original question, I think I now understand the point about image point versus focal point: the focal point is what you get with parallel rays, i.e. a source at infinity. If the source is close, so you don't have parallel rays, the focus moves correspondingly. The focus is literally where you see the image on a white paper, and the calculated focal length of 220 mm is irrelevant.)