Here’s the answer given in the journal:

No explanation is given. I had imagined something like a curved disk, but that wouldn’t give a circular plan view.

The nearest similar puzzle I know is the “architect’s puzzle,” in which you’re asked to imagine a single object that can cast three different shadows: a circle, a square, and a triangle. That’s possible, but it’s a distinctly different task, and it’s hard to see how to adapt that object into the one we’re seeking here.

09/27/2023 UPDATE: I’ve received a ton of mail about this puzzle — a million thanks to everyone who’s contributed; I’m constantly impressed by your intelligence, imagination, and resourcefulness. The consensus is that the solution was published upside down! And the puzzle as presented is a bit ambiguous. Reader Catalin Voinescu explains:

Note there are no additional lines to the semicircle that is the front elevation: that’s all there is to see from that direction. The plan view could be filled or not, and we can’t tell. So the figure must be the intersection of the bottom half of a horizontal cylinder surface (like a half-round gutter) and either a vertical cylinder surface, like a gutter downpipe, or a solid vertical cylinder. If you think of what a T joint of two cylinders of the same diameter looks like from the side, the joints appear as straight lines that bisect the angle between the cylinders. (They must be straight lines bisecting the angle for reasons of symmetry.)

So the given answer is one of two possibilities.

When the plan view is hollow, the figure is a closed curve in three dimensions: draw an ellipse with a long diameter times the short diameter, and bend the paper at 90 degrees along the short diameter. The top view is a circle, the side view a semicircle, the other side view is the angle made by the paper.

However, if the plan view was solid, the figure would be a curved surface: what you need to cut out of the half-round gutter to get the vertical cylinder to go through it. Viewed from the side, it would look like a triangle — like the given solution, but with a line at the bottom.

That second solution looks like this:

Here’s an interactive plot of the first solution, and here’s the second. Click and drag these figures to examine them. In both cases the side elevation is an inverted V, so the published solution is upside down (and I’m not crazy!).

My thanks to Drake Thomas and to Catalin for their work on all this, and thanks again to everyone else who wrote in about it. I’ve actually gone through the subsequent issues of the journal to see whether an erratum notice was ever published, and I don’t find one. I wonder how much mail this generated in 1958!