Step-by-step Derivation:

1. Understanding the Force: The two hemispherical shells are uniformly charged with the same type of charge (implied by a single $\sigma$). Due to electrostatic repulsion, they will tend to fly apart. The external force $F$ is applied to counteract this repulsion and hold them together. Therefore, we need to calculate the total electrostatic force exerted by one hemisphere on the other.

2. Electrostatic Pressure: A charged surface experiences an outward electrostatic pressure. The electric field just outside a uniformly charged spherical shell is $E_{out} = \frac{\sigma}{\varepsilon_0}$, and the field just inside is $E_{in} = 0$. The force on any small patch of charge $dq$ on the surface is due to the electric field produced by all other charges on the shell. This field, $E_{rest}$, can be shown to be the average of the fields just inside and outside, excluding the field of the patch itself. The field of the patch is $\frac{\sigma}{2\varepsilon_0}$ pointing outwards and inwards. The field from the rest of the sphere is $E_{rest}$ such that $E_{rest} + E_{patch} = E_{out}$ and $E_{rest} - E_{patch} = E_{in}=0$. This gives $E_{rest} = \frac{\sigma}{2\varepsilon_0}$.

3. Calculating Pressure: The force $dF$ on a small area element $dA$ with charge $dq = \sigma dA$ is given by $dF = dq \cdot E_{rest} = (\sigma dA) \left(\frac{\sigma}{2\varepsilon_0}\right)$. The electrostatic pressure $P$ (force per unit area) is therefore: $P = \frac{dF}{dA} = \frac{\sigma^2}{2\varepsilon_0}$ This pressure acts radially outwards at every point on the surface of the sphere.

4. Integrating the Pressure to Find Total Force: We need to find the net force exerted on one hemisphere by this outward pressure. Let's consider the top hemisphere. By symmetry, the horizontal components of the force will cancel out. We only need to sum (integrate) the vertical components of the force.

   Let's set up a coordinate system with the origin at the center of the sphere and the z-axis passing through the pole of the top hemisphere. The two hemispheres are separated by the xy-plane.

   Consider a small area element $dA$ on the hemisphere at an angle $\theta$ from the z-axis. The force on this element is $d\mathbf{F} = P \cdot dA \cdot \hat{r}$, where $\hat{r}$ is the radial unit vector.

   The component of this force along the z-axis is $dF_z = (P \cdot dA) \cos\theta$.

   The total force on the hemisphere along the z-axis is the integral of this component over the entire surface of the hemisphere: $F_z = \int_{hemisphere} P \cos\theta \, dA$

5. Projected Area Method: The integral $\int_{hemisphere} \cos\theta \, dA$ represents the projection of the hemispherical surface area onto the base plane (the xy-plane). This projected area is a circle of radius $R$. $A_{projected} = \pi R^2$

6. Final Calculation: The total repulsive force is the pressure multiplied by this projected area: $F_{repulsive} = P \times A_{projected} = \left( \frac{\sigma^2}{2\varepsilon_0} \right) (\pi R^2) = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$

   The force $F$ required to hold the hemispheres together must be equal in magnitude to this repulsive force: $F = \frac{\pi \sigma^2 R^2}{2\varepsilon_0}$

7. Determining Proportionality: From the final expression for $F$, we can see its dependence on $\sigma$ and $R$. The term $\frac{\pi}{2}$ is a dimensionless constant. $F \propto \frac{1}{\varepsilon_0} \sigma^2 R^2$

8. Comparing with Options: Comparing this proportionality with the given options: A: ${1 \over {{\varepsilon _0}}}{\sigma ^2}{R^2}$ - This matches our derived proportionality. B: ${1 \over {{\varepsilon _0}}}{\sigma ^2}R$ C: ${1 \over {{\varepsilon _0}}}{{{\sigma ^2}} \over R}$ D: ${1 \over {{\varepsilon _0}}}{{{\sigma ^2}} \over {{R^2}}}$

   Therefore, option A is the correct answer.