If your outer integral is over $z$ (you get to pick-the answer should come out the same) you need to figure out the range of $z$. In this case it runs from $0$ to $3$ so we have $\int_0^3 dz ($something). For the next integral, we get to consider $z$ to be fixed. If we do it over $y$, we need to figure out the range of $y$ at a given $z$. Now we have $\int_0^3 dz \int_{y_{min}}^{y_{max}} dy ($range of $x$ at this $(y,z))$. Now, considering $y$ and $z$ are fixed, you need to figure out the range of $x$, put that in the parentheses and you have your double integral.
Your setup is right. Here is the method you could have done to compute the volume.
Assume the density is $f(x,y,z) = 1$, so
$$V = \iiint_D \,dx\,dy\,dz$$
We are given that the solid is bounded by $z = x^2 + y^2 + 1$ and $z = 2 - x^2 - y^2$. As I commented under your question, you need to use cylindrical coordinates to evaluate the volume integral. Using the substitutions $x = r\cos(\theta)$, $y = r\sin(\theta)$ and $z = z$, we have $z = r^2 + 1$ and $z = 2 - r^2$. With some knowledge in graphs and functions, we see that the bounds are
$$\begin{aligned}
r^2 + 1 \leq z \leq 2 - r^2\\
0 \leq \theta \leq 2\pi\\
0 \leq r \leq \dfrac{1}{\sqrt{2}}
\end{aligned}$$
where $r = \frac{1}{\sqrt{2}}$ is found by solving for $r$ when $r^2 + 1 = 2 - r^2$. So for the volume triple integral, we have
$$\begin{aligned}
V &= \iiint_D \,dx\,dy\,dz\\
&= \iiint_D r\,dr\,d\theta\,dz\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2 + 1}^{2 - r^2}r\,dz\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(2 - r^2 - r^2 - 1)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(1 - 2r^2)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} (r - 2r^3)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\,d\theta\left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \cdot \dfrac{1}{8}\\
&= \dfrac{\pi}{4}
\end{aligned}$$
Best Answer
It looks like you set up the iterated integral correctly. $z= 1 - y^2$ and $z= y^2 - 1$ are the elliptic parabolas that extend out the $x$ axis, so $ 0\leq x\leq 2$ are the right bounds. Then you have $y$ between $-1$ and $1$ because of the intersection of $z = 1-y^2$ and $z = y^2 = 1$.
Doing this quickly, I also get $16/3$ as the answer.