So we all now agree on the specs of the bullet.
mass = 0.1kg
velocity =1000m/s
diameter of tip = 3.9mm or 0.0039m
length of bullet = 4in or 0.1016m
Let's calculate the force that would be exerted on Thor's head if it were to stop the bullet in less than equal to the length of the bullet (the stopping distance). This distance is 0.1016m as shown above.
Average stopping Force = Change in Kinetic Energy / stopping distance
= 1/2 x mass x velocity^2 / stopping distance
= 1/2 x (0.1kg) x (1000m/s)^2 / 0.1016m
= 492126 Newtons
Now 1 Newton = 0.22480894387096 pounds of force
So 492126N = 492126N x (0.22480894387096lb/1N) = 110634 lbs of force
2000 lbs of force = 1 ton of force
So 110634lbs = 110634lb x (1 ton/2000 lbs) = 55.3 tons of force
But the PEAK of this average stopping Force (more than 55.3 tons of force) is initially applied at the tip of the bullet the moment the bullet starts to deform.
Peak Pressure = Peak Force /Area of tip
> 492126N/ (pi x r^2]
= 492126N/[pi x (0.0039/2 m)^2]
= 4.12 x 10^10 Pascals
Now 1 pound per square foot (psi)= 6894.7572932 Pascals
So 4.12 x10^10 Pascals = (4.12 x 10^10 Pascals) x (1psi / 6894.7572932 Pascals)
= 5975004 psi
= 2987.5 tons per square inch (dividing by 2000 again)
This is the initial pressure. The pressure will decrease as the bullet deforms more.
Note: I assumed the bullet explodes on Thor and therefore deforms it entire length (4 in). If it richochets or mushrooms a little then the force is larger.
Also the actual tip is smaller, I used an overestimate by measuring the diameter below the tip, where it is wider. You will be surprised that significantly decreasing the diameter of the tip increases the pressure by a large amount.