$B$ is a first octant solid bounded by $y=x^2$, $x=y^2$, and $x+y-z=-5$
Since first octant, we know that $x,y,z\geq 0$
We know $y=x^2$, and therefore, $x=y^2\to x=x^4(\text{ from $y=x^2$)}\to x=0,1$
Since $0\leq x\leq 1$, this means $0\leq y\leq 1$, which implies that $5\leq z\leq 7$ (by plugging in $(x,y)=(0,0)\text { and } (1,1)$
So the volume is:
$$\int^{1}_{0}dx\int^{1}_{0}dy\int^{7}_{5}dz=2$$
Is this correct? Usually your second and last integrals are functions of the other, not constant values exactly so I'm not sure.
I tried to draw the solid:
Looking at the figure, it is easy to see that the integration will go from $0$ to $1$ with respect to $x$. If $x$ is given then the integration will go from $x^2$ to $\sqrt x$ with respect $y$. And, if $x$ and $y$ are given then the integration will go from $0$ to $x+y+5$. That is, the volume can be calculated as
$$\int_0^1 \int_{x^2}^{\sqrt x} \int_0^{x+y+5} 1\ dz\ dy \ dx.$$