A Banach space $Z$ is said to be injective if for for any bounded linear map $\varphi: X \rightarrow Z$ and for any Banach space $Y$ containing $X$ as a closed subspace, there exists a bounded linear extension $\tilde{\varphi}: Y \rightarrow Z $ such that $ \left\|\tilde{ \varphi }\right\| = \left\| \varphi\right\|$.
In the real case...the answer is yes. But in the complex case, $\mathbb{C}^2$, answer is No. How to establish this? Some examples?
The one-dimensional space $\mathbb R$ is injective, by the Hahn-Banach theorem.
If $X$ and $Y$ are injective, then $X\oplus_{\infty} Y$ is injective (subscript means using the norm $\|(x,y)\|_\infty = \max (\|x\|,\|y\|)$ on the direct sum.) This is because any bounded linear map into $X\oplus_{\infty} Y$ can be extended component-wise, and the definition of the norm is such that the norm of operator into $X\oplus_{\infty} Y$ is the maximum of the norms of its components.
Hence, $\ell_\infty^n$ is injective for every $n$ (so is $\ell_\infty$, but this is not needed here.) In particular, $\ell_\infty^2$ is injective. And $\ell_\infty^2$ is isometric to $\ell_1^2$: see here.