Hint: If $x \in [0, 2]$, then:
$$
y \leq \sqrt{4 - x^2} \iff y^2 \leq 4 - x^2 \iff x^2 + y^2 \leq 4 \iff r^2 \leq 4 \iff r \leq 2
$$
Note that in the $xy$-plane, $\{(x, y) \mid x \in [0, 2] \text{ and } y \in [0, \sqrt{4 - x^2}\}$ represents the Quadrant I section of a circle centred at the origin with radius $2$ (that is, it is the top-right quarter circle). So what should $\theta$ range between?
When switching from Cylindrical to Spherical, imagine taking a cross-section by slicing the region at an arbitrary angle using the half-plane $\theta$ to obtain an $rz$-plane. In this cross-section, we want:
$$
S = \{(r, z) \mid r \in [0, 2] \text{ and } z \in [0, \sqrt{16 - r^2}]\}
$$
That is, we are taking a quarter-circle of radius $4$ and making a vertical slice at $r = 2$ and keeping the left piece. Now because $(0, 4), (2, 0) \in S$, we are forced to let $\phi \in [0, \pi/2]$. As for $\rho$, it's a bit tricker as we transition from the arc $z = \sqrt{16 - r^2}$ to the line $r = 2$, which intersect at the threshold $(2, \sqrt{12})$. Notice that at this threshold, we have $\rho = 4$, so:
$$
2 = r = \rho\sin\phi = 4\sin\phi \implies \phi = \frac{\pi}{6}
$$
Hence, for $\phi \in [0, \pi/6]$, the outermost boundary for $\rho$ is:
$$
z = \sqrt{16 - r^2} \iff r^2 + z^2 = 4^2 \iff \rho = 4
$$
while for $\phi \in [\pi/6, \pi/2]$, the outermost boundary for $\rho$ is:
$$
r = 2 \iff \rho\sin\phi = 2 \iff \rho = 2\csc\phi
$$
Putting it all together, we obtain:
\begin{align*}
&\int_0^2 \int_0^\sqrt{4 - x^2} \int_0^\sqrt{16 - x^2 - y^2} \sqrt{x^2 + y^2} \, dz \, dy \, dx \\
&= \int_0^{\pi/2} \int_0^2 \int_0^\sqrt{16 - r^2} r (r \, dz \, dr \, d\theta) \\
&= \int_0^{\pi/2} \int_0^{\pi/2} \int_0^{\min\{4, 2\csc\phi\}} (\rho \sin \phi)(\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta) \\
&= \int_0^{\pi/2} \left[ \int_0^{\pi/6} \int_0^{4} \rho^3 \sin^2 \phi \, d\rho \, d\phi + \int_{\pi/6}^{\pi/2} \int_0^{2\csc\phi} \rho^3 \sin^2 \phi \, d\rho \, d\phi \right] d\theta
\end{align*}