We need to make certain assumptions:
the pipe is connected to the base of the tank;
the pipe is 12" in diameter;
the inclination of the pipe makes an immediate constant angle of 50° with the horizontal base of the tank and drops 100m vertically from the tank;
the friction of the pipe walls is negligible so that the water flow is unimpeded;
atmospheric pressure is assumed to be negligible compared with the water pressure;
the level of water in the tank is maintained at 20m (otherwise the flow rate will decrease as the tank empties).
[The volume of water in the tank is 150×150×20=450000m3. Density of water=1000kg/m3, so the mass of water in the tank, assuming the density of water is 1000kg/m3, is 4.5×108kg. The weight of water, assuming acceleration of gravity is 9.8m/s2, is 4.41×109N. I haven't used the weight of the water in the calculation, because, as you'll see, I don't think it's relevant to the solution.]
Assume the pressure acts on the circular cross-section of the pipe, which has a radius of r=6"=0.1524m, area=π(0.15242)=0.073m2.
We ignore the dimensions of the tank apart from the depth and just consider P=depth times density=20×1000×9.8=196000Pa. The force exerted by this pressure=196000×0.073=14301N (pressure times area). The length of the inclined pipe is 100/sin50=130.54m approx. The volume of water in the pipe when full is 100πr2=7.297m3, and its mass is 7297kg. We can use Newton's Law F=mass×acceleration to find the acceleration of water in the pipe, where F=14301+7297×9.8sin50=69082N. This force is partly due to the weight of water in the pipe, which is reduced by the inclination. So acceleration A is F/7297 m/s2=69082/7297=9.47m/s2 approx. We use the equation v2=2As where A is acceleration, v is velocity and s is distance (pipe length=130.54m). So v=√(2×9.47×130.54)=49.72m/s, the flow rate. Note that this figure doesn't take into account the effects of turbulence or friction which would tend to reduce the flow rate.
