Answer
$m=2k \pi$ and $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$
Work Step by Step
Here, $ds=\sqrt{(dx)^2+(dy)^2+(dz)^2}$
or, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$
Thus, $ds=\sqrt{(-2\sin t)^2+(2\cos t)^2}dt=2 dt$
Here, $m=\int_C k ds=2k\int_{-\pi/2}^{\pi/2} dt=2k \pi$
Now, $\overline {x}=\dfrac{1}{2k \pi}\int_{C} xk ds=\dfrac{1}{2 \pi}\int_{C} (2 \cos t) 2dt=\dfrac{4}{\pi}$
and $\overline {y}=\dfrac{1}{2k \pi}\int_{C} yk ds=\dfrac{1}{2 \pi}\int_{C} (2 \sin t) 2dt=0$
Hence, the mass of the wire is: $m=2k \pi$ and $(\overline {x}, \overline {y})=(\dfrac{4}{\pi},0)$