A correct solution came from Bradley University students Scott Small, Ross Pendergrast.  Other correct solutions were submitted by Francisco be Leon-Sotelo y Esteban, Bryan Fluhrer, Ahron Teitelman, Cee Ann Franklin, Juan Carlos Marivela, Larry Bell, Lisa Schechner, Paul Botham, Bill Webb, Dan Dima.

On the left is a sketch of the earth with center at O of radius r, with the International Space Station, S, at distance d from the center of the earth. Consider the earth's "horizon" as viewed by the astronauts on the ISS, that is, the points on the surface visible and furthest away from them.  These points lie on a circle on the earth's surface and the area visible to the astronauts lies "inside" this circle.  Let A be a point on the horizon, then the line AS is tangent to the earth at A and is perpendicular to the radius OA.  Note that |OA| = r, |OS| = d, so by similar triangles, |AB| = r²/d.  Since the earth is a surface of rotation, the surface area visible from the ISS is given by the standard integral , where f (x) = √(r² - x²) describes a great circle on the earth.  I admit this integral looks ugly, but it turns out to be absolutely trivial (try it, almost everything cancels!), and gives the visible area inside the horizon to be .  Of course, the total surface area of the earth is 4πr².  Put in the given data and you'll discover that the astronauts can see only about 2.67% of the surface at any one time! 

Several submissions assumed that earth inside the horizon forms a circle.  But this would be true only if the earth were flat!