@amistre64 I made the same confusion with the \(11\) and \(9\).
Either way, it's still believable that the center of mass of the system lies on the side of the arc. All of its mass is far off to the right. It's very dense, there. The mass of the lamina is uniform, so its extra 0.2kg (or around there if this is Earth (m = 2/g)) is spread closer to its base, not just on the edges where it would shift the lamina center of mass most leftward.
I wouldn't be able to guess, given the shapes and weights, but I would consider it a toss-up until the math was done.
@Vincent-Lyon.Fr found similar center of masses for each the lamina and arc as I did.
Instead of integrating with those values, I just took the general formulas from my book.
I'm going to use this coordinate system:|dw:1398456215345:dw|because we know that the center of mass for each, and thus the whole thing, will be on that axis \(x\) and having the \(y\)-axis where it is makes it convenient to find distance from \(O\) to the center of mass. And the \(y\)-axis there means I can use my formulas!! The lamina is in the negative \(x\) region and arc in the positive.
\({\sf \large Lamina}\\\qquad
x_{cm}=-\dfrac{4a}{3\pi}=-0.29708....{\rm\ m}\approx0.297{\rm\ m}\\
{\sf \large Arc}\\\qquad
x_{cm}=\dfrac{2a}{\pi}=0.44563....{\rm\ m}\approx0.446{\rm\ m}\)
So.....