The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV = B \iiint\limits_{\mathcal{B}} \, dV.$$
This is a paraboloid and its volume can be found using the cylindrical coordinate substition. We find $z = 2(x^2+y^2) = 2r^2$ and the limits for $z$ will be $2r^2 \leq z \leq \sqrt{c/2}$. This was found equating $z=2r^2 = c$, therefore $r = \sqrt{c/2}$.
$$
\begin{align}
\iiint\limits_{\mathcal{B}} \, dV & = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c r \, dz \, dr \, d \theta \\
& = 2 \pi \int_0^{\sqrt{c/2}} cr - 2r^3 \, dr \\
& = 2 \pi \left( \frac{cr^2}{2} - \frac{r^4}{2} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
& = \pi \left( c \cdot \frac{c}{2} - \frac{c^2}{4} \right) \\
& = \frac{c^2 \pi}{4}.
\end{align}
$$
Hence $M = (Bc^2 \pi)/4$.
Coordinates for center of mass are defined as
$$
\begin{align}
\overline{x} & = \frac{1}{M} \iiint x \rho \, dV \\
\overline{y} & = \frac{1}{M} \iiint y \rho \, dV \\
\overline{z} & = \frac{1}{M} \iiint z \rho \, dV.
\end{align}
$$
In physicist notation I've seen this written as
$$\mathbf{R} = \frac{1}{M} \iiint \rho \, \mathbf{r} \, dV.$$
It is interesting to note the following: the paraboloid is symmetric around $z$ axis. This means that the center of mass must be in the $z$ axis, for the $\overline{x}$ and $\overline{y}$ will cancel (if you don't believe this, write out the integral explicitly: you will have to integrate $\cos \theta$ and $\sin \theta$ over $[0,2 \pi]$, which is zero).
Therefore it is left for us to compute the $z$ coordinate. Leaving out the density out for a second (since it is uniform), we have
$$
\begin{align}
\iiint\limits_{\mathcal{B}} z \, dV & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c z r \, dz \, dr \, d \theta \\
& = 2 \pi \int_0^{\sqrt{c/2}} \frac{r}{2} \left( c^2 - 4r^4 \right) \, dr \\
& = \pi \int_0^{\sqrt{c/2}} c^2 r - 4r^5 \, dr \\
& = \pi \left( \frac{(cr)^2}{2} - \frac{2r^6}{3} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
& = \pi \left( \frac{c^2}{2} \cdot \frac{c}{2} - \frac{2}{3} \cdot \frac{c^3}{8} \right) \\
& = \pi \left( \frac{c^3}{4} - \frac{c^3}{12} \right) \\
& = \pi \left( \frac{c^3}{6} \right) \\
& = \frac{c^3 \pi}{6}.
\end{align}
$$
Finally
$$\overline{z} = \frac{B c^3 \pi}{6} \cdot \frac{4}{Bc^2 \pi} = \frac{2c}{3}.$$
Therefore
$$M = \frac{Bc^2 \pi}{4} \text{ and } \mathbf{R} = (\overline{x}, \overline{y}, \overline{z}) = \left( 0, 0, \frac{2c}{3} \right).$$
Hope this helps and best wishes.
$$\bar{\cos{\theta}} = \int dr \, r^2 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, \cos{\theta}$$
$$\bar{\sin{\phi}} = \int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \, \int d\theta \, $$
$$\bar{r} = \int dr \, r^3 f(r) \, \int d\phi \, \sin{\phi} \, \int d\theta \, $$
Then
$$\begin{align}\bar{r} \,\bar{\sin{\phi}}\,\bar{\cos{\theta}} &= \frac{\int dr \, r^3 f(r)\left (\int dr \, r^3 f(r) \right )^2 \int d\phi \, \sin^2{\phi} \left ( \int d\phi \, \sin{\phi} \right )^2\int d\theta \, \cos{\theta} \left ( \int d\theta \right )^2}{\left (\int dr \, r^2 f(r) \right )^3 \left ( \int d\phi \, \sin{\phi} \right )^3 \left ( \int d\theta \right )^3}\\ &= \frac{\int dr \, r^3 f(r) \, \int d\phi \, \sin^2{\phi} \,\, \int d\theta \, \cos{\theta} }{\int dr \, r^2 f(r)\, \int d\phi \, \sin{\phi}\, \int d\theta }\\ &= \bar{x}\end{align}$$
But I do not think you can work with the bare $\phi$ and $\theta$ because the trig functions introduce a nonlinearity that belies the linearity of the integrals.
Best Answer
Distance of any point from $z-$axis is given by $\sqrt{x^2+y^2}$, so density $\rho = \sqrt{x^2+y^2}$.
Note that the density $\rho$ is not a function of $z$ and hence the $z$ coordinate of center of mass will simply be $z = 1$.
You can also use the symmetry about $x-$axis to conclude that the $y$ coordinate of the center of mass will be $0$.
So you are left with finding $x$ coordinate for the center of mass. For that, consider mass of unit height (unit $z$) in $XY$ plane above $x-$axis (as there is symmetry above and below $x-$axis).
Mass $m = \displaystyle \int_{0}^{2} \int_0^3 \sqrt{x^2+y^2} \ dx \ dy$
And, $\overline{x} = \displaystyle \frac{1}{m} \int_{0}^{2} \int_0^3 x \sqrt{x^2+y^2} \ dx \ dy$
Just as a side note, the second integral can be evaluated over $dx$ using $u = y^2+x^2$ substitution. Then hyperbolic substitution works over $dy$.
To make it easier to evaluate the first integral - if you are familiar with line integral of vector fields and Green's theorem, please note line integral of the vector field $\frac{1}{3}(-y \sqrt{x^2+y^2}, x \sqrt{x^2+y^2})$ over the closed curve which is perimeter of the rectangle with vertices $A(0,0), B(3, 0), C(3,2), D(0,2)$, is same as the first integral to find mass. Line integral over $AB$ and $DA$ is zero. So after parametrization, this translates to line integral over $BC$ and $CD$ as below.
$\displaystyle \int_0^2 \sqrt{t^2+9} \ dt + \frac{2}{3}\int_0^3 \sqrt{t^2+4} \ dt$