Solution

The force on an infinitesimal area element of the shell is given by:
$$\sigma \, dS = dF$$
where $\sigma$ is the mechanical stress, and $dS$ is the differential surface area.

From geometry, the area of a spherical cap under a constant solid angle $\Omega$ is:
$$S = R^2 \Omega$$
Taking the differential:
$$dS = 2R \, dR \, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}$$

The gravitational force on a differential mass element is:
$$dF = g \, dm$$
By Gauss’s Law for gravity:
$$\oint \vec{g} \cdot d\vec{S} = -4\pi G M \quad \Rightarrow \quad g = \frac{GM}{R^2}$$

The mass of the differential shell element is:
$$dm = \rho \, dV$$

Assuming the shell is thin, the differential volume can be written in terms of $dS$:
$$dV = 4\pi R^2 \, dR = \frac{R \, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \, dS}{2}$$

Substitute into the expression for $dF$:
$$dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}$$

Now substitute into the original stress equation:
$$\sigma \, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2} \quad \Rightarrow \quad \sigma = \frac{GM \rho}{2R}$$

Substituting known values:

$$\sigma = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(4 \times 10^3)}{2 \times (1.496 \times 10^{11})} \approx \boxed{1.8 \times 10^{12} \, \text{Pa}}$$