What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration

due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the

acceleration due to gravity on the Earth's surface. That tells us that

the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.