= Solution
Choose the first $SU(2)$ as $SU_w(2)$ and embed $SU_c(3)$ in the upper-left three-by-three block of $SU(4)$. The <fundamental representations> then restrict as $\mathbf4\to\mathbf3\oplus\mathbf1$ and $\overline{\mathbf4}\to\overline{\mathbf3}\oplus\mathbf1$. Thus the <representation> in <4/c/solution>, forgetting <hypercharge>, restricts to
$$
(\mathbf2,\mathbf3)\oplus(\mathbf2,\mathbf1)
\oplus2(\mathbf1,\overline{\mathbf3})\oplus2(\mathbf1,\mathbf1).
$$
This contains every required $SU_w(2)\times SU_c(3)$ <representation>, with one extra singlet.
For the <hypercharge> embedding, let $X=\operatorname{diag}(1,1,1,-3)$ in the $SU(4)$ fundamental <representation>, and $T_{3R}=\operatorname{diag}(1/2,-1/2)$ in the second $SU(2)$. These commute with $SU_w(2)\times SU_c(3)$. The integer normalization in the paper is
$$
\boxed{Y=X+6T_{3R}.}
$$
Explicitly, this <circle group> embeds in the covering <group> by
$$
z\longmapsto\left(I_2,\begin{pmatrix}z^3&0\\0&z^{-3}\end{pmatrix},
\operatorname{diag}(z,z,z,z^{-3})\right).
$$
Both nontrivial blocks have determinant one, and the map is injective since the first three entries of the $SU(4)$ block recover $z$.
On $(\mathbf2,\mathbf1,\mathbf4)$, $T_{3R}=0$, so the <branching of a representation> gives $(\mathbf2,\mathbf3)_1\oplus(\mathbf2,\mathbf1)_{-3}$. On $(\mathbf1,\mathbf2,\overline{\mathbf4})$, the <U(1) charges> under $X$ are $-1$ on $\overline{\mathbf3}$ and $3$ on $\mathbf1$, while $6T_{3R}$ takes values $\pm3$. Hence
$$
\begin{aligned}
(\mathbf1,\mathbf2,\overline{\mathbf4})&\longrightarrow
(\mathbf1,\overline{\mathbf3})_{-1+3}\oplus
(\mathbf1,\overline{\mathbf3})_{-1-3}\\
&\qquad\oplus(\mathbf1,\mathbf1)_{3+3}\oplus(\mathbf1,\mathbf1)_{3-3}.
\end{aligned}
$$
Combining the two summands gives
$$
\boxed{R\longrightarrow R_F\oplus(\mathbf1,\mathbf1)_0.}
$$
This is the <Pati-Salam charge embedding>, with $X=3(B-L)$. In terms of the common <Standard Model> convention $Q_{\rm electric}=T_{3L}+Y_{\rm SM}$, the paper's integer <hypercharge> is $Y=6Y_{\rm SM}$.
The extra neutral singlet can be interpreted as the left-handed charge conjugate $\nu_R^c$ of a <right-handed neutrino>. It is a <sterile neutrino> under the <Standard Model> <gauge group>, and completes the sixteen <Weyl fermions> of one family in this <half-spin representation>. Its neutrality permits a <Majorana mass term> without violating the remaining <Standard Model> <gauge group>.
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