Step 1: Understanding the Concept:
In a balanced Wheatstone bridge, the ratio of resistances in adjacent arms is equal: $\frac{P}{Q} = \frac{R}{S}$.
Step 2: Detailed Explanation:
Assuming the bridge layout follows the standard ratio (where $2/X = 6/12$):
$$\frac{2}{X} = \frac{6}{12}$$
$$\frac{2}{X} = \frac{1}{2}$$
$$X = 4\text{ }\Omega$$
Step 3: Final Answer:
The value of $X$ is 4 $\Omega$.