= Solution
Use the question's indexing convention: $m_{\lambda,\mu}$ is the <multiplicity> of the <irreducible> <Specht module> $S^\mu_{\mathbb C}$ in the <permutation module> $M^\lambda_{\mathbb C}$. In notation for semistandard-tableau <Kostka numbers> this is often written $K_{\mu\lambda}$, with the indices reversed.
Let $S_\lambda=S_{\lambda_1}\times S_{\lambda_2}\times\cdots$ be the <Young subgroup>. The <permutation character> is
$$
\xi^\lambda=\operatorname{Ind}_{S_\lambda}^{S_n}\mathbf1,
$$
and $\xi^\lambda(g)$ counts the $\lambda$-tabloids fixed by $g$. Equivalently it counts <set partition>[ordered partitions] into labelled rows of the prescribed sizes, each row <invariant> under $g$. <Character> inner products and <Frobenius reciprocity> give the purely character-theoretic computation
$$
\boxed{m_{\lambda,\mu}=\langle\xi^\lambda,\chi_\mu\rangle_{S_n}
=\frac1{|S_\lambda|}\sum_{h\in S_\lambda}\chi_\mu(h).}
$$
The <characters> of <symmetric groups> are real-valued, so no conjugation is needed in the displayed <subgroup> average.
For $\lambda=(2,1^3)$, the <Young subgroup> has two elements, the identity and one <transposition>. Thus $m_{(2,1^3),\mu}=(\chi_\mu(1)+\chi_\mu((12)))/2$, and part (a) gives
$$
\boxed{\begin{array}{c|rrrrrrr}
\mu&(5)&(4,1)&(3,2)&(3,1,1)&(2,2,1)&(2,1^3)&(1^5)\\\hline
m_{(2,1^3),\mu}&1&3&3&3&2&1&0
\end{array}}
$$
The resulting <dimension> is $1+3\cdot4+3\cdot5+3\cdot6+2\cdot5+4=60=5!/2$, the <dimension> of the <permutation module>.
This is the <character average for Kostka numbers>.
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