Answer: $x_c=18.2$ and $y_c=9.5$.
Who can solve with explanation?
Given the distributive property of the centroid, calling $A_i, i=1,2,3$ the areas of the triangle, rectangle and circle, respectively, and $P_i=(x_i,y_i)$ the respective centroids, we have
\begin{align} x_c&=\frac{A_1x_1+A_2x_2-A_3x_3}{A_1+A_2-A_3}\\ &=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot2a-\pi r^2\cdot2a}{a^2+4a^2-\pi r^2} \end{align}
and similarly for $y_c$
\begin{align} y_c&=\frac{A_1y_1+A_2y_2-A_3y_3}{A_1+A_2-A_3}\\ &=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot a-\pi r^2\cdot a}{a^2+4a^2-\pi r^2} \end{align}
given the centroid of the triangle $(2a/3, 2a/3)$ and the common centroid of the rectangle and of the circle $(2a,a)$.
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Given the distributive property of the centroid, calling $A_i, i=1,2,3$ the areas of the triangle, rectangle and circle, respectively, and $P_i=(x_i,y_i)$ the respective centroids, we have
\begin{align} x_c&=\frac{A_1x_1+A_2x_2-A_3x_3}{A_1+A_2-A_3}\\ &=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot2a-\pi r^2\cdot2a}{a^2+4a^2-\pi r^2} \end{align}
and similarly for $y_c$
\begin{align} y_c&=\frac{A_1y_1+A_2y_2-A_3y_3}{A_1+A_2-A_3}\\ &=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot a-\pi r^2\cdot a}{a^2+4a^2-\pi r^2} \end{align}
given the centroid of the triangle $(2a/3, 2a/3)$ and the common centroid of the rectangle and of the circle $(2a,a)$.