To overcome hill resistance, you need 2977 Nm or 2195 ft*lbs of torque at the wheels.
Assumed that the gravitional acceleration in your situation is $9.81m/s^2$, as it approximtely is everywhere on Earth, your weight is:
The hill resistance which drags you backwards is :
To keep any given speed, the force must be the same at the outer side of your wheel. So any speed that is achieved, is maintained. (theoretically, no friction included)
Assumed you meant that the diameter(not radius) of your wheels is 830mm,
this translates in a torque at the wheels of
To drive your vehicle to the speed of 27.78m/s up a hill of 7 degrees, you need
$P=F*v=7173 * 27.78 = 199.257 kW$ (or roughly 270hp)
of power, when there wouldn't be any friction.
But there will be friction, so you'll need more power and torque. This isn't calculable though, since there isn't enough information supplied in the question. At a normal asphalt road with an expectable loaded truck of 6 tonnes, i'd guess that you need roughly 50% more power, and therefore 50% more torque at the same speed. You also need a margin to accelerate to that speed in not-too-much time. NB: this is just a wild guess, don't rely on it for any important decisions.
Don't forget that the transmission in a vehicle will lower the torque demand at the motor, at the price of the demand for a higher angular speed of the engine, relative to that of the wheels.