To see what the statement means, please look at Derived Categories, Definition \ref{derived-definition-generators}. Please keep in mind that since $R$ is Noetherian and regular objects of $D(R)$ are perfect if and only if they are bounded and have finite cohomology modules, see Lemma \ref{lemma-regular-perfect} \medskip\noindent Let $d = \dim(R)$. In the next two paragraphs we first handle the cases $d = 0, 1$ but we suggest the reader skip these. \medskip\noindent If $d = 0$, then $R$ is a finite product of fields. Then every $R$-module is projective. Thus every object of $D(R)$ is isomorphic to a direct sum of its cohomologies and every $R$-module is a summand of a free $R$-module, so $D_{perf}(R) = \langle R \rangle_1$ and the lemma is true. \medskip\noindent If $d = 1$, then every object of $D^b(R)$ is isomorphic to the direct sum of its cohomology objects as follows from Derived Categories, Lemma \ref{derived-lemma-ext-2-zero} and Algebra, Lemmas \ref{algebra-lemma-finite-gl-dim-finite-dim-regular} and \ref{algebra-lemma-projective-dimension-ext}. Hence to prove $D_{perf}(R)$ is strongly generated by $R$ it suffices to prove that $M \in \langle R \rangle_2$ for every finite $R$-module $M$. This is clear from the fact that $M$ has a finite projective resolution of length $1$. \medskip\noindent Assume $d \geq 2$. We will prove that $D_{perf}(R) \subset \langle R \rangle_{2(d^2 - 1)}$ which finishes the proof (we make no attempt to get a good bound). In the argument we will use both canonical and stupid truncations, see Homology, Section \ref{homology-section-truncations} for our conventions. Let $M$ be an object of $D_{perf}(R)$. Choose $a \leq b$ such that $H^i(M) = 0$ for $i \not \in [a, b]$. For $k \in 2(d - 1)\mathbf{Z} \cap [a, b + 2d]$ choose a quasi-isomorphism $$ P_k^\bullet \to \tau_{\leq k}M $$ where $P_k^\bullet$ is a complex of finite free $R$-modules with $P_k^i = 0$ for $i \geq k + 1$. Then we consider the map $$ N_k = \sigma_{\geq k - d + 2}P_k^\bullet \to P_k^\bullet \to \tau_{\leq k}M \to M $$ Clearly the maps $H^i(N_k) \to H^i(M)$ are isomorphisms for $i = k - d + 3, \ldots, k$ (no cases for $d = 2$) and surjective for $i = k - d + 2$. Observe that $H^i(N_k) = 0$ for $i \not \in [k - d + 2, k]$. Choose a distinguished triangle $$ \bigoplus\nolimits_{k \in 2(d - 1)\mathbf{Z} \cap [a, b + 2d]} N_k \to M \to L \to \left( \bigoplus\nolimits_{k \in 2(d - 1)\mathbf{Z} \cap [a, b + 2d]} N_k \right)[1] $$ Examining the long exact sequence of cohomology we find that $H^i(L) = 0$ for $i \in [k - d + 2, k]$ for $k \in 2(d - 1)\mathbf{Z} \cap [a, b + 2d]$. By Lemma \ref{lemma-split-complex-regular} this means that $L$ splits as a direct sum $$ L = \bigoplus\nolimits_{k \in 2(d - 1)\mathbf{Z}} L_k $$ with $H^i(L_k)$ nonzero only for $i \in [k - 2(d - 1) + 1, k - (d - 1)]$ and $L_k$ zero for almost all $k$. Details omitted. Each $L_k$ is a perfect complex. By the result of the next paragraph we see that $\bigoplus N_k$ and $L = \bigoplus L_k$ are in $\langle R \rangle_{d^2 - 1}$. Hence $M$ is in $\langle R \rangle_{2(d^2 - 1)}$ by Derived Categories, Lemma \ref{derived-lemma-generated-by-E-explicit} as desired. \medskip\noindent Let $K$ be an object of $D_{perf}(R)$ with $H^i(K)$ nonzero only for $i = 0, \ldots, d - 2$. Then we claim $K$ is in $\langle R \rangle_{d^2 - 1}$. First, using $d - 2$ distinguished triangles of canonical truncations (Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}) we reduce to proving that any finite $R$-module $M$ is in $\langle R \rangle_{d + 1}$. This is clear because $M$ has a resolution of length $d$ by finite projective $R$-modules. See Derived Categories, Lemmas \ref{derived-lemma-generated-by-E-explicit} and \ref{derived-lemma-in-cone-n}.