You mention point mass. Yes, the volume also matters. If your second black box contains the same mass but over a bigger volume, then the spacetime curvature it will cause will be less extreme than the black hole in the first box. The book I most like on this topic is Kip Thorne's Black Holes and Time Warps. IMO Thorne is a better explainer than Hawking.
I'm pretty sure by the time you're outside the box, assuming it's the same size for both, you can't tell anymore. I'm quite confident this is the case for classical gravity and a spherically symmetric "box", and I don't think tides or relativistic corrections are noticeably different far away from the horizon. (Yeah, you'll feel the black hole's tides, but stars have tides too.)
I'm not a relativity expert, but couldn't you tell the difference between a point mass (or just significantly smaller volume) and a star in this black box scenario? At a great distance, they would appear the same gravitationally, but as you get closer, the star would appear less massive. Since more mass would be pulling on you at an angle, rather than directly toward the center.
Definitely not for classical gravity. https://en.wikipedia.org/wiki/Shell_theorem Point 1 there says that the physical extent of the mass doesn't matter for a spherically symmetric object, to which a star is close enough.
Are we talking about the volume of the event horizon? If I understood it correctly, the total of the mass of a black hole is in its singularity. The volume of the event horizon will depend on the total mass of the black hole.
Oh I just mean when comparing (A) block hole in a black box, vs (B) a non-black-hole start of the same mass, B will likely be over a large volume, and hence will produce different spacetime curvature.
