= Solution
There is one complex <irreducible representation> for each <conjugacy class>, hence three. The <trivial representation> $\mathbf1$ and <sign representation> $\varepsilon$ are distinct and one-dimensional. The <sum of squares of irreducible representation dimensions> leaves
$$
1^2+1^2+d^2=6,\qquad\boxed{d=2.}
$$
Call the remaining <irreducible representation> $\mathbf2$. The <permutation representation> on $\mathbb C^3$ has <character> equal to the number of fixed objects, namely $(3,1,0)$ on the ordered <conjugacy classes> $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector <trivial representation> gives $(2,0,-1)$, whose <character inner product> with itself is one. This constructs $\mathbf2$ and gives the <character table>
$$
\boxed{\begin{array}{c|rrr}
\text{class}&e&(12)&(123)\\
\text{class size}&1&3&2\\\hline
\chi_{\mathbf1}&1&1&1\\
\chi_{\varepsilon}&1&-1&1\\
\chi_{\mathbf2}&2&0&-1
\end{array}.}
$$
To obtain every <Clebsch-Gordan decomposition of group representations>, multiply the <characters> pointwise. In particular, $\chi_\varepsilon^2=\chi_{\mathbf1}$, $\chi_\varepsilon\chi_{\mathbf2}=\chi_{\mathbf2}$ and
$$
\chi_{\mathbf2}^2=(4,0,1)=\chi_{\mathbf1}+\chi_\varepsilon+\chi_{\mathbf2}.
$$
Thus all ordered pairs of <irreducible representations> are covered by
$$
\boxed{\begin{array}{c|ccc}
\otimes&\mathbf1&\varepsilon&\mathbf2\\\hline
\mathbf1&\mathbf1&\varepsilon&\mathbf2\\
\varepsilon&\varepsilon&\mathbf1&\mathbf2\\
\mathbf2&\mathbf2&\mathbf2&\mathbf1\oplus\varepsilon\oplus\mathbf2
\end{array}.}
$$
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