Step 1: Understanding the Concept
Sonometer frequency is \(f\propto\dfrac{1}{L}\), so the 1.0 m wire gives a higher frequency than the 1.06 m wire.
Step 2: Set up
Let the fork have frequency \(n\). Then \(f_{1.0}=n+6\) and \(f_{1.06}=n-6\).
Step 3: Use inverse proportion
\(\dfrac{n+6}{n-6}=\dfrac{1.06}{1.0}\), so \(n+6=1.06n-6.36\).
Step 4: Calculation
\(0.06n=12.36\), so \(n=206\ \text{Hz}\).
Final Answer:
The frequency of the fork is 206 Hz.
\[ \boxed{\text{(B) }206\text{ Hz}} \]