Instrument aperture sizes on Hubble Telescope
EDIT: You can't have a very wide field, and simultaneously a very high resolving power, in an instrument with a large focal ratio (or any instrument, really), unless you use sensors with an unrealistic pixel count. Details below:

The linear size of the focal plane is large - one document says it's the size of a "dinner plate". But the instrument is an f/24, which is a large focal ratio. At that ratio, a big linear size in the focal plane doesn't mean much in terms of angular size.

From what I've seen, the Wide Field Camera has a total field of 4.7 arcmin, or 280 arcsec according to this document:

http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf

Anyway, yes, that's a small field, if you compare it with ground-based amateur astrographs, which can have a field as wide as several degrees of arc.

But the Hubble telescope has a very high resolving power - 0.05 arcsec. Let's take your information and assume the WFC has a total field of 3 arcmin.

3 arcmin / 0.05 arcsec = 3600 pixels in theory, as a linear size of the WFC

If you wanted the WFC to have a bigger total field, you'd have to either sacrifice resolving power, or use a sensor with an enormous pixel count, which is not feasible.

The WFC uses several sensors, each one of them 800 x 800 pixels.



TLDR: Wide field, or great resolving power - pick one. They chose resolving power, and they compensate the narrow field by doing multiple passes.