I claim in general, the area of the triangle formed by the medians of triangle ABC is $\frac{3}4[ABC]$. To see, let $D,E,F$ be the midpoints of the respective sides as shown in the diagram. Then construct point $G$ such that $BF$ is parallel and equal to $DG$. Then $DFGC$ is a parallelogram. Let the intersection of the diagonals be $H$. Then $H$ is the midpoint of $CF$ and $DG$. Since $BEGC$ is a parallelogram, $\angle{AEG}=\angle{ABC}$, $BC=EG$ and $AE=EB$. Thus triangles $AEG$ and $EBC$ are congruent, and so $AG=EC$. Therefore triangle $ADG$ has side lengths of the medians of $ABC$. Now we can easily compute $\frac{[ADG]}{[ABC]}=\frac{2[ADH]}{2[ADC]}=\frac{AH}{AC}=\frac{3}4$, and so therefore $[ADG]=\frac{3}4[ABC]$. In this case, we have the answer to be $92\cdot\frac{3}4=\boxed{69}$.