How do you solve, and what is the solution to the following question:

A sattelite of Mars has a period of 459 minutes. The mass of mars is
6.42x10^23 kg. Determine the radius of the sattelite's orbit.

I have that 
v=c/t
v=(2piR)\(27540 sec)
v= (pi R)\(13770)

then, I plugged that into:
(mv^2)/R=(GmM)\R^2

but im not getting anything useful out of that.

A satellite in a circular orbit of radius r with angular velocity w
experiences an acceleration of w^2*r.  This acceleration must be
provided by the gravity of the planet and is equal to G*M/r^2, where G
is the universal gravitational constant, M is the mass of the planet,
and r is the distance from the center of the planet to the satellite's
orbit.

Equate these two expressions:

w^2*r = G*M/r^2

solve for r.

Note that the angular velocity is expressed in radians/time.  One
complete orbit is 2*pi radians, so the angular velocity of your
satellite is 2*pi/459*minute = 2*pi/(459*60)second).

The individual who commented is exactly right, of course, and the
answer with the data given is 9.35 x 10^7m.  The published data gives
9380km for the mean radial distance of Phobos, and it appears that the
difference is in the period, which is listed as 459.5 minutes.

This is my first time on this site, and I don't know the etiquette,
but it appears to me that the original question asked for
understanding of the solution method, so in the hopes that this is of
some use, let me try.

If the orbit is treated as a circle (and "the" radius doesn't have
much meaning if you don't), then if you take an instantaneous look at
the system, there is a radius vector of length R directed from the
center of Mars out to the center of Phobos and a "circular" or
tangential velocity v at right angles to the radius vector.  If you
look at the system after some interval t, the radius vector has
rotated through some angle and the displacement vector can be resolved
into two vectors, one tangential and one radial.  The tangential
vector is of length vt, and the radial vector is of length
s=(1/2)*at^2.  You can now draw a right triangle whose hypotenuse is
R,and with one side of length vt and the other side of length (R-s). 
Using the Pythagorean theorem,

R^2 = (vt)^2 + (R-s)^2

which, if solved for R, gives

R = (v^2 * t^2 + s^2)/(2*s) = (4 * v^2 + a^2 * t^2) / (4 * a)

If you let the time interval between peeks go to zero, the second term
in the numerator drops out and the equation reduces to

R = v^2 / a

The questioner already knows what v is.  It is the circumference of
the orbit divided by the period, or

v = 2 * pi * R / T

where T is the orbital period.  The questioner used the equation for
the force of gravitation, rather than the equation for he acceleration
due to gravity, so dropping the additional mass m

F = ma = GmM/R^2

gives 

a = GM/R^2

and plugging the eequations for v and a into the equation for R,
collecting all  of the terms and solving for R, gives

R^3 = T^2 * G * M / (4 * pi^2)