Aharon,
This sounds like something a professor would dream up. Here's how I would approach this problem:
Let's assign some values:
rho = density of air at STP = 1.2 kg/m3
mu = viscosity of air at STP = 1.8e-5 Ns/m2
V = velocity of vehicle = 33.33 m/s (120 km/hr)
L = length of flag = 0.25 m
b = width of flag = 0.04 m
t = thickness of flag = 0.002 m
First calculate the Reynolds number, ReL = rho*V*L/mu = 5.6e5. This Reynolds number is close to the transition from laminar flow to turbulent flow for a flat plate. Since this problem is much easier tackled assuming laminar flow, that's what I would do.
Now, there are two sources of drag. The primary source is stagnation of the flow at the leading edge of the blunt flag. For a thin plate, the coefficient of drag, C, is 2.0. Drag is calculated by D = C*0.5*rho*V2*A, where A is the frontal area (t*b). I get a drag force of 0.11 N.
The other source of drag is due to the boundary layer on either side of the plate. Assuming laminar flow and a Blasius-type boundary layer, the coefficient of drag can be estimated by C = 1.328/Re0.5 (square root of Reynolds number) = 1.8e-3. Drag is now calculated by D = C*0.5*rho*V2*A, where A is the flag area (L*b). I get a drag force of 0.012 N on one side, making the drag force due to both boundary layers 0.024 N.
This gives a total drag force of 0.13 N. I would assume this is acting half way up the flag, or 0.48 m from the base of the flag pole, which gives a moment of 0.062 Nm.
Remember, there are a couple of important assumptions here. The first is an assumption of laminar flow, which is borderline at this Re. The other is that the flow that stagnates at the leading edge of the flag does not disturb the formation of a Blasius-type boundary layer along the flag's length. Probably not the case, but a relatively accurate assumption (I think) none-the-less.
Hope this helps,
Haf