At the suface to be 1/10 of what is now?

The acceleration due to gravity is not dependent on the speed of rotation of the earth. However, it does affect the force you perceive, as the centripetal acceleration counteracts the force of gravity. The centripetal acceleration is v^2/r, where r is the radius of the earth. Set it equal to 9/10 the value of g. This will give you a value for v that satisfies the requirement at the equator.

Approx. V = 16 kmph. A lot higher peripheral speed than I expected. But Thanks.

Uhhh... I think your calculation is a little bit off. Like three orders of magnitude or more.

Yea. The units are off. Sqrt(0.9 * 9.8 m/s^2 * 6,437,376 m) = sqrt(5.67777e7 m^2/s^2) = 7535.1 m/s = 16,855.5386 mph