= Solution
Put $\alpha=e^{2\pi i/3}$, so $\alpha^3=1$ and $1+\alpha+\alpha^2=0$. Since the <group> is <Abelian>, every <conjugacy class> is a singleton. The <character table> is
$$
\boxed{\begin{array}{c|ccc}
&0&1&2\\\hline
\chi_0&1&1&1\\
\chi_1&1&\alpha&\alpha^2\\
\chi_2&1&\alpha^2&\alpha
\end{array}.}
$$
For the <character inner product>, complex conjugation gives
$$
\langle\chi_j,\chi_k\rangle
=\frac13\sum_{g=0}^2\overline{\chi_j(g)}\chi_k(g)
=\frac13\left(1+\alpha^{k-j}+\alpha^{2(k-j)}\right).
$$
For $j=k$, all three summands are one, giving unit norm. For $k-j\equiv1\pmod3$ the sum is $1+\alpha+\alpha^2=0$; for $k-j\equiv2\pmod3$ it is $1+\alpha^2+\alpha=0$. This checks every distinct pair, proving
$$
\boxed{\langle\chi_j,\chi_k\rangle=\delta_{jk}.}
$$
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