1. In Young's double slit experiment, the intensity at a point with phase difference $\phi$ is $I = I_0 \cos^2\left(\frac{\phi}{2}\right),$ where $I_0$ is the maximum intensity.

2. The path difference is given as $\Delta x = \frac{\lambda}{6}.$ The corresponding phase difference is $\phi = \frac{2\pi}{\lambda}\,\Delta x = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{6} = \frac{\pi}{3}.$

3. Substitute into the intensity formula: $\frac{I}{I_0} = \cos^2\left(\frac{\phi}{2}\right) = \cos^2\left(\frac{\pi}{6}\right).$ Now, $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.$ So, $\frac{I}{I_0} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.$

4. Check options:

• A: $\frac{3}{4}$ ✅
• B: $\frac{1}{\sqrt{2}}$ ❌
• C: $\frac{\sqrt{3}}{2}$ ❌
• D: $\frac{1}{2}$ ❌

Hence, the correct answer is A.