Suppose $\mathcal{B}$ is a Banach space which is not isomorphic to a Hilbert space. And $\mathcal{M}$ is a closed subspace of $\mathcal{B}$. What condition should I pose on $\mathcal{B}$ such that the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$?

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The comment of M.González looks really good to me. But that kind of space may be too stringent and hard to check for me. So I am considering to loosen the closed subspace to finite dimensional subspace, that is, suppose $\mathcal{M}$ is a finite dimensional subspace of $\mathcal{B}$. What condition should I pose on $\mathcal{B}$ such that the quotient space $\mathcal{B}/\mathcal{M}$ is isomorphically isometric to some closed subspace of $\mathcal{B}$? Since finite dimensional subspace is always complemented, there exists a closed subspace $\mathcal{N}$ such that $$\mathcal{B}=\mathcal{M}\oplus\mathcal{N}.$$ And then we can prove $\mathcal{B}/\mathcal{M}$ with natural quotient topology is isomorphic to $\mathcal{N}$ where natural quotient topology is defined as $$ \|x+\mathcal{M}\|_{\mathcal{B}/\mathcal{M}}:=\inf_{y\in\mathcal{M}}\|x-y\|_{\mathcal{B}},\quad\forall x\in\mathcal{B} $$ But it seems not isometric in general. I am trying to put some semi-inner-product in the sense of Lumer-Giles on $\mathcal{B}$. There is a theorem (Theorem 71 of Semi-Inner Products and Applications by S. S. Dragomir) which says if $\mathcal{B}$ is strictly convex, reflective, smooth and has an semi-inner-product (L-G), then there is a unique decomposition of $\mathcal{B}$ as $$ \mathcal{B}=\mathcal{M}\oplus\mathcal{M^{\perp}} $$ where $\mathcal{M^{\perp}}$ is the orthogonal complement of $\mathcal{M}$ in the sense of Lumer-Giles. Clearly, here $\mathcal{M}^{\perp}=\mathcal{N}$. But I cannot prove $$\inf_{y\in\mathcal{M}}\|x-y\|_{\mathcal{B}}=\|P_{\mathcal{M}^\perp}(x)\|_{\mathcal{M^{\perp}}}$$ where $P_{\mathcal{M}^\perp}(x)$ is the projection of $x\in\mathcal{B}$ on $\mathcal{M}^\perp$. Once we prove that, the isometry will be true? Any remedy for this? Or any other condition should be posed on $\mathcal{B}$? Thanks a lot!