Answer
$$2 \pi \sqrt {13} (k+3 \pi)$$
Work Step by Step
Since, $ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dy}{dt})^2} dt$
We have: $x=2\cos t \implies \dfrac{dx}{dt}=-2 \sin t$
and $y=2 \sin t \implies \dfrac{dy}{dt}= 2 \cos t$ and $z=3 \implies \dfrac{dz}{dt}=3$
So, $P(x,y) =ds=\sqrt { (-2 \sin t)^2+ (2 \cos t)^2 +(3)^2}dt= \sqrt {13} \ dt$
We have: $\rho=k+z=k+3t$
So, the mass the wire can be computed as:
$$\int_C \rho \ ds= \sqrt {13} \int_{0}^{2 \pi} k+3t \ dt$$
So, $$\int_C \rho \ ds=\sqrt {13} \int_{0}^{2 \pi} k+3t \ dt \\=\sqrt {13} [kt+\dfrac{3t^2}{2}]_0^{2 \pi} \\= 2 \pi \sqrt {13} (k+3 \pi)$$
