Solution

The force on a small area of a sphere can be defined as the mechanical stress multiplied by the area of this area:
$$σdS=dF$$
From stereometry we know that the area of a sphere is determined by the product of the solid angle and the square of the radius of this sphere:
$$S=R²Ω$$
The thickness of the sphere is unknown to us, although it is not required. Assuming the solid angle of the sphere to be constant, we take the derivative of the area:
$$dS=2RdRΩ$$

$$dR=\frac{dS}{2RΩ}$$
Force on a small fraction of the sphere's mass:
$$dF=g*dm$$
Applying Gauss's theorem, we obtain the flow of the gravitational field through the area of the sphere:
$$\oint gdS=-4πGM$$

$$-4πR²g=-4πGM$$
Let's find the gravitational field strength:
$$g=\frac{GM}{R²}$$
Mass of a small part of the sphere:
$$dm=ρdV$$
The derivative of volume, which can be written through the derivative of area:
$$dV=4πR²dR=\frac{RdS}{2}$$
Let's substitute the force into the very first equation we wrote down:
$$σdS=\frac{GM}{R²}ρ\frac{RdS}{2}$$
From here we get the working formula for mechanical stress:
$$σ=\frac{GMρ}{2R}$$
Substituting the numbers, we get:
$$\boxed{σ=1.8*10¹²Pa}$$