Evaluate the integral from the inside out and watch the surfaces appear. The innermost integral is \[\int_0^{2-x} dz=2-x.\] This is exactly the height under the plane $z=2-x$ above the base point $(x,y)$, so the solid is capped above by this plane and below by $z=0$.
What remains is \[\int_0^2\int_0^{4-y^2}(2-x)\,dy\,dx,\] a double integral of the height function $2-x$ over a base region in the xy-plane. That base region is tied to the curve $x=4-y^2$, a parabola in the xy-plane, which as a boundary in three-dimensional space (with no z restriction) is a parabolic cylinder.
So the full triple integral is the volume swept out under the plane $z=2-x$ over the base bounded by the parabolic cylinder $x=4-y^2$, for $0\le x\le 2$. Since $z=2-x\ge0$ throughout this range, the solid stays entirely above the xy-plane, and the description matching this shape among the choices is the region bounded by the parabolic cylinder $x=4-y^2$ and the plane $z=2-x$. \[\boxed{\text{(B) The region bounded by the parabolic cylinder } x=4-y^2 \text{ and the plane } z=2-x}\]