This is slightly more difficult than it seems. it depends on how you define "how far". There are at least 2 ways of defining the radius-- you can define it as the circumference of a circle which stays at a constant distance from the centre of the earth, or you can define it as the distance from the centre of the earth-- they are different, and even the change in radius is different. So let me take the circumference definition, the drop for small distance l is about 3gl^2/c^2 (g is 10m/s^2,c^2 is 10^17m^2/s^2) so in 1 km the drop would be about 3x10^(-10)m. This against a rise of l^2/2r which over that 1km would be about 0.1m.
[Editor: So to put it simply, if you shine a beam of laser light horizontally, parallel to the surface of the Earth, after 1km it will be "pulled" down by gravity 3x10^(-10)m, or 0.0000000003m. However, due to the curvature of the Earth's surface, the light beam would actually appear to rise about 0.1m over that 1km distance.]