In the classical case we choose $R = \Lambda$. More generally, if the residue field extension $k/k'$ is separable, then there exists a unique finite \'etale extension $\Lambda^\wedge \to R$ (Algebra, Lemmas \ref{algebra-lemma-complete-henselian} and \ref{algebra-lemma-henselian-cat-finite-etale}) of the completion $\Lambda^\wedge$ of $\Lambda$ inducing the extension $k/k'$ on residue fields. \medskip\noindent In the general case we proceed as follows. Choose a smooth $\Lambda$-algebra $P$ and a $\Lambda$-algebra surjection $P \to k$. (For example, let $P$ be a polynomial algebra.) Denote $\mathfrak m_P$ the kernel of $P \to k$. The Jacobi-Zariski sequence, see (\ref{equation-sequence-extended}) and  Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}, is an exact sequence $$ 0 \to H_1(\NL_{k/\Lambda}) \to \mathfrak m_P/\mathfrak m_P^2 \to \Omega_{P/\Lambda} \otimes_P k \to \Omega_{k/\Lambda} \to 0 $$ We have the $0$ on the left because $P/k$ is smooth, hence $\NL_{P/\Lambda}$ is quasi-isomorphic to a finite projective module placed in degree $0$, hence $H_1(\NL_{P/\Lambda} \otimes_P k) = 0$. Suppose $f \in \mathfrak m_P$ maps to a nonzero element of $\Omega_{P/\Lambda} \otimes_P k$. Setting $P' = P/(f)$ we have a $\Lambda$-algebra surjection $P' \to k$. Observe that $P'$ is smooth at $\mathfrak m_{P'}$: this follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-slice-smooth-given-element}. Thus after replacing $P$ by a principal localization of $P'$, we see that $\dim(\mathfrak m_P/\mathfrak m_P^2)$ decreases. Repeating finitely many times, we may assume the map $\mathfrak m_P/\mathfrak m_P^2 \to \Omega_{P/\Lambda} \otimes_P k$ is zero so that the exact sequence breaks into isomorphisms $H_1(L_{k/\Lambda}) = \mathfrak m_P/\mathfrak m_P^2$ and $\Omega_{P/\Lambda} \otimes_P k = \Omega_{k/\Lambda}$. \medskip\noindent Let $R$ be the $\mathfrak m_P$-adic completion of $P$. Then $R$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Namely, it is a complete local Noetherian ring (see Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian}) and its residue field is identified with $k$. We claim that $R$ works. \medskip\noindent First observe that the map $P \to R$ induces isomorphisms $\mathfrak m_P/\mathfrak m_P^2 = \mathfrak m_R/\mathfrak m_R^2$ and $\Omega_{P/\Lambda} \otimes_P k = \Omega_{R/\Lambda} \otimes_R k$. This is true because both $\mathfrak m_P/\mathfrak m_P^2$ and $\Omega_{P/\Lambda} \otimes_P k$ only depend on the $\Lambda$-algebra $P/\mathfrak m_P^2$, see Algebra, Lemma \ref{algebra-lemma-differential-mod-power-ideal}, the same holds for $R$ and we have $P/\mathfrak m_P^2 = R/\mathfrak m_R^2$. Using the functoriality of the Jacobi-Zariski sequence (\ref{equation-sequence-functorial}) we deduce that $H_1(L_{k/\Lambda}) = \mathfrak m_R/\mathfrak m_R^2$ and $\Omega_{R/\Lambda} \otimes_R k = \Omega_{k/\Lambda}$ as the same is true for $P$. \medskip\noindent Finally, since $\Lambda \to P$ is smooth we see that $\Lambda \to P$ is formally smooth by Algebra, Proposition \ref{algebra-proposition-smooth-formally-smooth}. Then $\Lambda \to P$ is formally smooth for the $\mathfrak m_P$-adic topology by More on Algebra, Lemma \ref{more-algebra-lemma-formally-smooth}. This property is inherited by the completion $R$ by More on Algebra, Lemma \ref{more-algebra-lemma-formally-smooth-completion} and the proof is complete. In fact, it turns out that whenever $\underline{R}|_{\mathcal{C}_\Lambda}$ is smooth, then $R$ is isomorphic to a completion of a smooth algebra over $\Lambda$, but we won't use this.