Answer
The mass of the lamina is
$$
\begin{aligned}
m &= \pi k.
\end{aligned}
$$
The center of mass is located at the point
$$
(\bar{x}, \bar{y})=\left(0, \frac{3}{ \pi}\right).
$$
Work Step by Step
The boundary of a lamina consists of the semicircles $y=\sqrt {1-x^{2}}$ and $y=\sqrt {4-x^{2}}$ together with the portions of the x-axis that joins them, if the density at any point is inversely proportional to its distance from the origin.
Then the distance from a point $(x,y)$ to the center of the origin is $\frac{1}{\sqrt {x^{2}+y^{2}}}$. Therefore the density function is
$$
\rho(x, y)=\frac{k}{\sqrt {x^{2}+y^{2}}}
$$
where $k$ is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then $\sqrt {x^{2}+y^{2}}=r $ and the region $D$ is given by:
$$ D=\left\{(r, \theta) | 1 \leq r \leq 2, \quad 0 \leq \theta \leq \pi \right\} .$$
The mass of the lamina is
$$
\begin{aligned}
m &=\iint_{D} \rho(x, y) d A \\
&=\iint_{D} \frac{k}{\sqrt {x^{2}+y^{2}}} d A \\
&=\int_{0}^{\pi} \int_{1}^{2} \frac{k}{r} \cdot r d r d \theta \\
&=k \int_{0}^{\pi} d \theta \int_{1}^{2} d r \\
&=k(\pi)\left[ r\right]_{1}^{2} \\
&= \pi k.
\end{aligned}
$$
Now, we find:
$$
\begin{aligned}
M_{y} &=\iint_{D} x \rho(x, y) d A \\
&=\int_{0}^{\pi} \int_{1}^{2}(r \cos \theta)(k / r) r d r d \theta \\
&=k \int_{0}^{\pi} \cos \theta d \theta \int_{1}^{2} r d r \\
&=k[\sin \theta]_{0}^{\pi}\left[\frac{1}{2} r^{2}\right]_{1}^{2} \\
&=k(0)\left(\frac{3}{2}\right)=0
\end{aligned}
$$
and
$$
\begin{aligned}
\\
M_{x} &=\iint_{D} y \rho(x, y) d A \\
&=\int_{0}^{\pi} \int_{1}^{2}(r \sin \theta)(k / r) r d r d \theta \\
&=k \int_{0}^{\pi} \sin \theta d \theta \int_{1}^{2} r d r \\
&=k[-\cos \theta]_{0}^{\pi}\left[\frac{1}{2} r^{2}\right]_{1}^{2} \\
&=k(1+1)\left(\frac{3}{2}\right)\\
&=3 k
\end{aligned}
$$
Therefore, the center of mass is located at the point
$$
(\bar{x}, \bar{y})=\left(0, \frac{3k}{ \pi k }\right)=\left(0, \frac{3}{ \pi}\right).
$$