1. Interpret the figure

   A disc of radius $a$ has a smaller circular hole of radius $\dfrac{a}{2}$ cut out from it.

   From the standard figure for this problem, the smaller hole is internally tangent to the bigger disc and lies along the radius through $O$. Hence:

   • Center of the big disc is at $O$.
   • Radius of big disc $= a$.
   • Radius of hole $= \dfrac{a}{2}$.
   • Distance of the hole's center from $O$ is $a-\frac{a}{2}=\frac{a}{2}.$

   Let the centroid of the remaining part lie on the horizontal line through $O$, on the side opposite to the hole.

2. Use the method of subtraction of masses

   Treat the full disc as a positive mass and the removed smaller disc as a negative mass.

   Since the lamina is uniform, mass is proportional to area.

   • Mass of full disc: $M = \sigma \pi a^2$
   • Mass of removed disc: $m = \sigma \pi \left(\frac{a}{2}\right)^2 = \sigma \pi \frac{a^2}{4} = \frac{M}{4}$

3. Take origin at $O$

   Let the center of the removed hole be at $x=\frac{a}{2}.$ The center of the full disc is at $x=0.$

   Then the $x$-coordinate of centroid of remaining part is $x_{cm} = \frac{M(0)-m\left(\frac{a}{2}\right)}{M-m}.$

4. Substitute $m=\dfrac{M}{4}$

   $x_{cm} = \frac{0-\frac{M}{4}\cdot \frac{a}{2}}{M-\frac{M}{4}}$

   $x_{cm} = \frac{-\frac{Ma}{8}}{\frac{3M}{4}}$

   $x_{cm} = -\frac{Ma}{8}\cdot \frac{4}{3M}$

   $x_{cm} = -\frac{a}{6}.$

   The negative sign means the centroid is on the side opposite to the hole.

   Therefore, its distance from point $O$ is $\boxed{\frac{a}{6}}.$

5. Check options

   • A: $\dfrac{1}{6}a$ ✅
   • B: $\dfrac{2}{3}a$ ❌
   • C: $\dfrac{5}{6}a$ ❌
   • D: $\dfrac{10}{11}a$ ❌

So the correct option is A.