a space shuttle is in a cic ular orbit at a height of 250 km, where the acceleration os earth,s gravity is 93 percent of its surface value. that is the period of its orbit

The distance from the center of the earth is increased by a factor 1/sqrt(0.93) = 1.0370. That is redundant information; they already tell you the altitude. The numbers are consistent.

You could use Kepler's third law for the period. I assume that your "that" is supposed to be "What"

You can also get the answer by requiring that the centripetal acceleration be 0.93 g, and solve for V.

V^2/R = 0.93 g

R = 250 + Re = ?
(Look up the radius of the earth, Re)

The period P is then given by
V * P = 2 pi R