Only 24g, and don't worry about temperature or radiation
Gravity is proportional to the square of distance.
At 10 AU, gravity will be 100 times smaller than normal gravity from the sun.
Studies have estimated the super-heavy black hole in the milky way center has a mass of 3.7 million or 4.1 million solar masses. As a result, you'll get 4e6/100 = 40,000 times Solar gravity at the point where you teleport your person.
On Earth, at the distance we orbit the sun, the gravitational pull of the sun is only 0.0006 of the strength of the earth's gravity on the surface of the earth. So when teleported at 10 AU from the galaxy center we'd experience 0.0006 x 40,000 is a force of 24g. In free fall, you won't notice..
Temperature and radiation may be a much bigger issue.. near a black hole, you'll have 10 million degrees.
However, at 10Au you'll be far away from the Schwartzshield radius of the black hole:
"Located 26,000 light-years from the Sun, our galaxy’s central black hole, Sagittarius A(star), has a radius about 17 times that of the Sun, meaning that it would sit well within Mercury’s orbit."
.. a distance of 10AU, far away from the Schwartzshield radius, will reduce the heat considerably.
Other potential issues, e.g. the Hawking radiation emitted,
This Hawking radiation is very low energy
An evaporating black hole would be detectable from Earth only if it went off within the solar system, or at best no further away than the nearest star.