Solution:

There are many solutions to this problem, which is straightforward. The given triangle is a right $3$-$4$-$5$ triangle, so the circumcenter is the midpoint of the hypotenuse. Coordinatizing for convenience, put the vertex at $(0,0)$ and the other vertices at $(0,18)$ and $(24,0)$. Then the circumcenter is $(12,9)$. The centroid is at one-third the sum of the three vertices, which is $(8,6)$. Finally, since the area equals the inradius times half the perimeter, we can see that the inradius is $(18 \cdot 24 / 2) /([18+24+30] / 2)=6$. So the incenter of the triangle is $(6,6)$. So the small triangle has a base of length $2$ and a height of $3$, hence its area is $3$.