Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating. Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (α), change in temperature ($\Delta T$), and the original dimensions of the bar.

First, compute the linear expansion of the bar if it were free to expand. The change in length ($\Delta L$) due to thermal expansion can be computed through the formula:

$$ \Delta L = \alpha L_0 \Delta T $$

Given:

• $$ \alpha = 10^{-5} { }^{\circ} C^{-1} $$
• $ L_0 = 1 \text{ m} $
• $$ \Delta T = 100^{\circ} C $$

Thus:

$$ \Delta L = 10^{-5} \times 1 \times 100 = 0.001 \text{ m} $$

This is the change in length that the bar would undergo if not constrained.

However, in this scenario, the bar is constrained and does not actually expand. This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain:

$ \sigma = E \epsilon $

Where the strain ($\epsilon$) under constrained conditions due to thermal expansion is:

$$ \epsilon = \frac{\Delta L}{L_0} = \frac{0.001}{1} = 0.001 $$

Therefore:

$$ \sigma = 0.5 \times 10^{11} \times 0.001 = 5 \times 10^7 \mathrm{ N/m}^2 $$

This stress is the force per unit area. To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar:

$$ F = \sigma A = 5 \times 10^7 \mathrm{ N/m}^2 \times 10^{-3} \mathrm{ m}^2 = 5 \times 10^4 \mathrm{ N} $$

Thus, the compressive force developed in the bar is $$50 \times 10^3 \mathrm{~N}$$.

Hence, the correct answer is Option B:

$$50 \times 10^3 \mathrm{~N}$$