r = 0.2667 m
P = 2.068427E+8 Pa
Now, we can solve and find out why this is a major problem. First, find the area of the pipe's exit point, assuming that it's a straight cut. This is 6th grade math.
A = π * r^2
A = π * (0.2667 m)^2
A = π * (0.7113 m^2)
A = 0.2235 m^2
Now using the given pressure and the calculated area, find the force of the oil coming out of the pipe.
F = P * A
F = (2.068427E+8 Pa) * (0.2235 m^2)
F = 46,220,607 N
This gusher is basically like a rocket engine flipped over. 46 million newtons is a full order of magnitude larger than the thrust of the early rockets that NASA sent into space. So you have to overcome that force. Doesn't matter which way you try to do it, you have to overcome this force, and that's obviously a huge problem.
To try to put things in perspective, look at it this way. If we were just to drop a single, solid block of mass
m = F/ a
m = (46,220,617 N) / (9.81 m/s^2)
m = 4,711,582 kg
I imagine you want that in Imperial units, which comes to about 10.4 million pounds. In other words, 5,200 tons (short). So what does 5,200 tons (short) "look like"? It's the equivalent mass of *seventy-seven* of the US Army's front-line M1A2 Abrams main battle tanks fully outfitted for combat with fuel and ammunition. Then you have to have them all stacked together in a single mass, because if you tried to just plop them down one after another, they'd just be blown aside by the oil "rocket."