Solution:

Lemma: If a triangle $ABC$ lies in a rectangle $KLMN$ with sides $KL = a$ and $LM = b$, then the area of the triangle is less than or equal to $\frac{ab}{2}$.

Proof of the lemma: Without any loss of generality, assume that among the distances of $A, B, C$ from $KL$, that of $A$ is between the other two. Let $\ell$ be the line through $A$ and parallel to $KL$. Let $D$ be the intersection of $\ell$ and $BC$, and $x, y$ the distances of $B, C$ from $\ell$ respectively. Then the area of $ABC$ equals $\frac{AD(x+y)}{2} \leq \frac{ab}{2}$, since $AD \leq a$ and $x + y \leq b$, and we are done.

Now back to our problem. Let us cover the circle with $24$ squares of side $\frac{1}{6}$ and $8$ other irregular and equal figures as shown in Figure 9, with boundary consisting of an arc on the circle and three line segments. Call $S = ADNM$ one of these figures. One of the line segments in the boundary of $S$ is of length $AD = AB - DB = \sqrt{AC^2 - BC^2} - \frac{2}{6} = \sqrt{\left(\frac{1}{2}\right)^2 - \left(\frac{1}{6}\right)^2} - \frac{1}{3} = \frac{\sqrt{2} - 1}{3}$.

The boundary segment $MN$ goes through the center $C$ of the circle, forming with the horizontal lines an angle of $45^\circ$. The point in $S$ with maximum distance from the boundary segment $AB$ is the endpoint $M$ of the arc on the boundary of $S$. This distance equals $ME = MF - EF$ (since $CMF$ is isosceles), $= \frac{\sqrt{2}}{2} CM - \frac{1}{6} = \frac{\sqrt{2}}{4} - \frac{1}{6} = \frac{3\sqrt{2} - 2}{12}$.

So $S$ can be put inside a rectangle $R$ with sides parallel to $AD, ND$ of lengths $\frac{\sqrt{2} - 1}{3}$ and $\frac{3\sqrt{2} - 2}{12}$. So the triangle formed by any three points inside this figure has an area less or equal to $\frac{1}{2} \cdot \frac{\sqrt{2} - 1}{3} \cdot \frac{3\sqrt{2} - 2}{12} = \frac{8 - 5\sqrt{2}}{72} < \frac{1}{72}$.

Also, the triangle formed by any three points inside any square of side $\frac{1}{6}$ has an area less or equal to $\frac{1}{2} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{72}$.

By the Pigeonhole Principle, we know that among the $65$ given points there exist $3$ inside the same one of the $32$ squares and irregular figures of the picture covering the given circle. Then according to the above, the triangle formed by these $3$ points has an area not exceeding $\frac{1}{72}$ as wanted.

Figure 9: Exercise C3.