What are additional external forces that keep the car uniformly moving forward? Given a problem as follows.

A torque of 500 Nm applied to axle of the wheel of a car moving uniformly. Find the forward friction exerted by the ground on the tire. Given that the wheel radius is 0.25 m.

My calculation: The total torque must be zero, we have
\begin{align}
F\times 0.25 &= 500\\
F&=2000 \; \text{N}
\end{align}
Question
In my understanding, friction is the only external force acting on the car. How can the car keep moving uniformly? What are the additional external forces that oppose the friction to keep the  car moving uniformly forward?
 A: 
In my understanding, friction is the only external force acting on the
car. How can the car keep moving uniformly? What are the additional
external forces that oppose the friction to keep the car moving
uniformly forward?

The external forces opposing the motion of the car include air resistance and tire rolling resistance. In order for the car to keep moving at constant velocity, the external 2000 N static friction force that the ground exerts forward on the car needs to equal the sum of all the external forces acting in the opposite direction to the the motion of the car, for a net force of zero.

What is the tire rolling resistance? It is different from =2000
newton I calculated?

Yes it is different.
The force you calculated is the static friction force acting forward (horizontally) on the car.
Rolling resistance is due to the inelastic deformation that the rubber of the tire experiences when it is compressed while contact with the road. It results in heating. It is one of the reasons fuel economy is less when tires are under inflated. More rubber is inelastically deformed when a tire is under inflated.

Sorry, is there kinetic friction between tires and ground?

No. Moreover, kinetic friction is not the same thing as rolling resistance. Kinetic friction is the friction that results when a car skids.
Kinetic friction occurs when the force exerted by the tire backward on the road reaches the maximum possible static friction force $F_{s}(max)$ acting forward on the car, where
$$F_{s}(max)=\mu_{s}N$$
Where $\mu_s$ is the coefficient of static friction and $N$ is the vertical reaction force of the road on the tire. It will equal the fraction of the weight of the car supported by the tire.
As an example, a compact car of mass 1300 kg weighs about 12,740 N. If the tire supports 1/4 of the weight, then $N$=3,185 N. If the coefficient of static friction is 0.7 then the maximum possible static friction force would be 9,229 N. Since this greater than the actual static friction force of 2000 N given in your example, the tire would not skid, and therefore there would be no kinetic friction.
Hope this helps.
