= Solution
Apply the <Murnaghan–Nakayama rule> successively to the $w$ disjoint $e$-cycles. A complete term requires a sequence of $w$ removable $e$-hooks. If $w>w_e(\lambda)$, no such sequence exists after the <Weight of a partition> is exhausted, so the character value is zero.
Suppose $w=w_e(\lambda)$. Every complete sequence ends at the <Core of a partition> $C_e(\lambda)$. Under the <abacus divisible-hook correspondence>, a removal chooses one cell from one component $\lambda^{(i)}$ of the <quotient of a partition>. The choices of which runner is used occur in
$$
\binom{w}{|\lambda^{(0)}|,\ldots,|\lambda^{(e-1)}|}
$$
orders. Within runner $i$, the signed complete removal sum is the degree $\chi^{\lambda^{(i)}}(1)$, and all inter-runner removal orders have the common <Sign of an abacus hook-removal sequence> $\varepsilon\in\{1,-1\}$. The remaining permutation $\gamma$ acts on the core, giving
$$
\chi^\lambda(\rho\gamma)
=\varepsilon
\binom{w}{|\lambda^{(0)}|,\ldots,|\lambda^{(e-1)}|}
\chi^{C_e(\lambda)}(\gamma)
\prod_{i=0}^{e-1}\chi^{\lambda^{(i)}}(1).
$$
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