Answer
$\Big(\frac{-0.8R}{3},0\Big)$
Work Step by Step
Lets take C to be located at the origin.
C=(0, 0)
C'=(0.8R, 0)
Let's name the given shape as A, the smaller circle B, and the larger circle C.
density$=\rho$
thickness$=t$
$m_C=4\rho t\pi R^2$
$m_B=\rho t\pi R^2$
$m_A=3\rho t\pi R^2$
$x_Cm_C=x_Am_A+x_Bm_B$
$(0)(4\rho t\pi R^2)=(0.8R)(3\rho t\pi R^2)+x_B(\rho t\pi R^2)$
$x_B=\frac{-0.8R}{3}$
$y_Cm_C=y_Am_A+y_Bm_B$
The shape is symmetric over the y axis so all values of y are equal to 0.
Thus, the coordinates of the center of mass is $\Big(\frac{-0.8R}{3},0\Big)$
