= Solution
One left-handed <Standard Model> family requires the <SU(5)> <representation> content $\mathbf{10}\oplus\overline{\mathbf5}$. No single <irreducible representation> in <4/b/solution> contains both. At the level of containing the required fields, the smallest choice is
$$
\boxed{\mathbf{15}\oplus\overline{\mathbf6}\downarrow
=(\mathbf{10}\oplus\overline{\mathbf5})\oplus\mathbf5\oplus\mathbf1.}
$$
The additional <SU(5)> multiplets are a $\mathbf5$ and a <trivial representation>. Another minimal choice using only two of the listed <irreducible representations> is
$$
\mathbf{15}\oplus\overline{\mathbf{15}}\downarrow
=(\mathbf{10}\oplus\overline{\mathbf5})\oplus(\overline{\mathbf{10}}\oplus\mathbf5).
$$
Its additional multiplets form the <conjugate representation> of a family. More generally, for multiplicities $(m_6,m_{\bar6},m_{15},m_{\overline{15}})$, <branching of a representation> gives
$$
(m_6+m_{15})\mathbf5\oplus(m_{\bar6}+m_{\overline{15}})\overline{\mathbf5}
\oplus m_{15}\mathbf{10}\oplus m_{\overline{15}}\overline{\mathbf{10}}
\oplus(m_6+m_{\bar6})\mathbf1.
$$
Thus containing a family requires $m_{15}\geq1$ and $m_{\bar6}+m_{\overline{15}}\geq1$.
For a consistent chiral <grand unified theory>, containment alone is insufficient. Normalize the cubic <gauge anomaly> coefficient by $A(\mathbf6)=1$. For the <special unitary group> $SU(N)$, $A(\Lambda^2\mathbf N)=N-4$, so for <SU(6)> it is $2$. To verify this directly, take a traceless diagonal generator with entries $x_i$: its <exterior square> has entries $x_i+x_j$, and
$$
\sum_{i<j}(x_i+x_j)^3=(N-1)\sum_i x_i^3+3\sum_{i\ne j}x_i^2x_j
=(N-4)\sum_i x_i^3.
$$
The <conjugate representation> reverses this cubic <gauge anomaly>. The combination $\mathbf{15}\oplus\overline{\mathbf6}$ has coefficient $2-1=1$, so cannot alone be a quantum-consistent chiral <SU(6)> <gauge theory>. The minimal anomaly-free choice with one net <chiral gauge spectrum> of a family is instead
$$
\boxed{\mathbf{15}\oplus2\overline{\mathbf6}\downarrow
=(\mathbf{10}\oplus\overline{\mathbf5})\oplus(\mathbf5\oplus\overline{\mathbf5})\oplus2\mathbf1.}
$$
The additional nonsinglet fields form a <vectorlike gauge spectrum> that can acquire an <SU(5)>-invariant mass, and the two <gauge singlets> may serve as sterile <neutrinos>. More generally, precisely one net <chiral gauge spectrum> of a family with vectorlike extra nonsinglets requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar6}-m_6=2$; these conditions also cancel the cubic <SU(6)> <gauge anomaly>. The $\mathbf{15}\oplus\overline{\mathbf{15}}$ choice above is anomaly-free but has zero net <chirality>.
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