I guess the answer is $2$, but I cannot prove it.
I mean supremum of the minimum $B/A$ among all $A$.
Notice:
Central convex curve means the convex curve is symmetric about a point in it, that is, when you rotate the curve 180 degrees about the point, it comes back to the original curve.
For the triangle, which is considered the most asymmetric, $2$ is a lower bound. That is why I think $2$ is a lower bound for all other convex curves.
I need to calculate a minimum $r=B/A$ such that for all of the closed convex curves，there always exists a centrally symmetric convex curve (with area $A\cdot r$) circumscribed about it. A lot of thanks.
I understand your question to be asking about the supremum over all planar convex sets $A$ of the ratio $\text{Area}(B)/\text{Area}(A)$, where $B$ is chosen to be a centrally-symmetric curve containing $A$ of minimal area.
This supremum is $2$. To attain this bound, take $A$ to be any triangle; any possible $B$ must contain the inversion of $A$ about its center, so $B$ has area at least that of the best convex hull of $A$ and a translation of $-A$. But for a triangle, such convex hulls can be pretty easily analyzed to see that their area is always at least twice that of the starting triangle. (Hint: fix a side of the triangle, and consider the area of $B$ as a function of the altitude between the side of $A$ and the corresponding side of $-A$.)
To see that we can obtain a ratio of at most $2$ for any convex set $A$, inscribe $A$ in a rectangle with one side parallel to its diameter (i.e., the line between two maximally-distant points on $A$). By considering the points where $A$ touches the sides of this rectangle and the fact that it is convex, we can see that $A$ contains a quadrilateral with area exactly half that of the rectangle, so the rectangle is of area at most twice that of $A$. But rectangles are centrally symmetric, so we are done.