First, there is no power if there is no motion. So, you need to find the «starting» torque required first.
Second, before knowing what is required, you have to determine the maximum traction force your truck can support.
Based on
this, a RWD 2-axle truck backing uphill would be similar to a FWD going uphill. Therefore, the maximum grade the truck can handle (not even moving, just holding) is:
tan\theta = \mu\frac{\frac{l_f}{L}}{1 + \mu\frac{h}{L}}
Where \frac{l_f}{L} is the portion of the truck's weight on the rear axle, \mu is the
tire-ground friction coefficient (typical: 0.8) and h & L are the height of the center of gravity and wheelbase of the truck, respectively.
If the truck begins to move, then you must add the inertia (\lambda ma) and the rolling resistance, which modify the previous equation like this:
\frac{sin\theta + \lambda\frac{a}{g} + f_r}{cos\theta} = \mu\frac{\frac{l_f}{L}}{1 + \mu\frac{h}{L}}
Where \frac{a}{g} is the acceleration in g's (typical: 0.2-0.3), \lambda is the
mass factor to take into account the rotational inertia (not sure what it would be for a truck, but typical for a passenger car in first gear is about 1.2) and f_r is the
rolling resistance coefficient (typical: 0.008).
If you find out that you can still go uphill, the traction force (F_r) needed at the rear axle would be:
F_r = mg\left(sin\theta + \lambda\frac{a}{g} + f_r\right)
Where mg is the total weight of the truck.
The engine torque needed is found knowing the rear tire radius, r_r, the total gear ratio between the engine and the tires, GR, and the transmission efficiency, \eta (typical: 0.85-0.9):
T_e = \frac{F_r r_r}{\eta GR}
Once the truck starts to move, the power required
at the axle is based on its velocity v:
P = F_r v
At the engine, the power will be:
P_e = \frac{P}{\eta (1-i)}
Where i is the transmission slip (clutch, tire) which is usually around 0.02-0.05.
