These are some crude calculations to get a lower limit, so I’ll outline the process for this. This is also just so that you’d see something was there, not necessarily that you’d be able to tell what it was. I’m going to ignore the atmosphere isn’t making it harder to see things, which isn’t really the case, but simplifies the math. The largest telescope (A) is likely going to be limited by the atmosphere to about the same seeing as the 10 inch telescope since the atmosphere is an additional constraint. So for A, this is more what you’d get if a telescope that big was in space just above the earth’s atmosphere.

I’m using a distance to the moon of 365,000 km, and presuming that we’re looking at blue light with a wavelength of 500 nm. The diameter I use for each of those is as follows:
A. Gran Telescopio CANARIAS, 10.4 meters
B. 10 inches is a good average size (a 10 inch dobsonian is about $600)
C. 7×50 binoculars, 50 mm
D. decently dilated, 8 mm

The angular resolution of a telescope, at it’s best, is going to be the wavelength divided by the diameter. Once we have the resolution, then the resolution multiplied by the distance to the moon will give us the size of object that it’d be able to resolve.

Here’s the numbers I get from some really quick math:
A. 20m
B. 730m
C. 4 km
D. 23 km

Keep in mind that this is the shortest dimension, so this is just a measurement of minimum girth. And if you really want it identifiable, you’d want to go bigger than this. It also won’t work if it’s just the outline, you’d need to fill the whole thing in.

For reference, the lunar module for Apollo 11 was about 10m across.
And for Hubble with a 2.5m mirror, it’d be about 75 meters.

(Hopefully no dumb errors on the math)