= Solution
Choose a maximal <Brauer pair> $(D,b_D)$ for $B$, and let $N_G(D,b_D)$ be the subgroup normalizing $D$ and stabilizing the local block $b_D$. The <inertial quotient of a block> and its order, the <inertial index of a block>, are
$$
E(B)=N_G(D,b_D)/(DC_G(D)),\qquad e=|E(B)|.
$$
The <Brauer extended first main theorem> says that maximal local blocks of defect $D$ are parametrized by the normalizer orbits of defect-$D$ blocks of $k(DC_G(D))$, with stabilizer quotient of order prime to $p$. For abelian $D$, conjugation embeds $E(B)$ in $\operatorname{Aut}(D)$. When $D=C_{p^n}$, its prime-to-$p$ subgroups are cyclic of order dividing $p-1$, so $e\mid p-1$.
We use two precise structural results for a <block with cyclic defect group>, with $k$ an <algebraically closed field>. First, the <basic algebra> is a <Brauer tree algebra> with $e$ edges and exceptional multiplicity $m=(p^n-1)/e$; its simple modules are indexed by its edges. Second, its <stable Auslander–Reiten quiver> has shape
$$
\mathbb ZA_{p^n-1}/\langle\tau^e\rangle.
$$
The vertices are the isomorphism classes of nonprojective indecomposable modules. These are the cyclic-defect structure and indecomposable-classification results, rather than a claim that all Brauer trees are Morita equivalent to a star. The tube description is also recorded in https://arxiv.org/pdf/1908.07833[Hiss–Lassueur, Section 1].
The first result gives $l(B)=e$. In the second, there are $p^n-1$ positions in each slice and $e$ slices modulo the translation. Thus there are $e(p^n-1)$ nonprojective indecomposables. A finite-dimensional split <algebra> has one <principal indecomposable module> that is a <projective cover> for each <simple module>, so there are another $e$ projective indecomposables. Therefore
$$
\boxed{l(B)=e,\qquad\#\operatorname{ind}(B)=e(p^n-1)+e=p^n e.}
$$
The total includes projectives; it is different from the count of ordinary irreducible characters. This is the <indecomposable count for a cyclic defect block>.
Now let $p=2$ and $D=C_2\times C_2$, the <Klein four-group>. The extended theorem makes $E(B)$ a subgroup of order prime to $2$ of $\operatorname{Aut}(D)\cong\operatorname{GL}_2(\mathbb F_2)\cong S_3$. Its order divides $6$ and is odd, so
$$
\boxed{e=1\text{ or }3.}
$$
For the remaining conclusion we use the central-defect quotient fact: if $Z\le Z(L)$ is a central $p$-subgroup, blocks of $kL$ correspond to blocks of $k(L/Z)$ under the quotient, and their defect groups quotient by $Z$. Central block idempotents lift uniquely here because the kernel is generated by central nilpotent elements; an off-diagonal corner lies in its own product with that nilpotent ideal and is therefore zero. Defect groups quotient by $Z$ by the relative-trace, or equivalently Brauer-pair, characterization. In particular a block with central defect group $D$ has a defect-zero block as its quotient by $J(kD)$.
First suppose $D\le Z(G)$, and write $A=B$. Since $k$ is an <algebraically closed field>, the defect-zero quotient is $\overline A\cong\operatorname{Mat}_r(k)$ for some $r$. The <algebra> $A$ is a direct summand of $kG$ as a $kD$-module, hence is projective and therefore free over the <local ring> $kD$. Its rank is $r^2$, since reduction modulo $J(kD)$ gives the $r^2$-dimensional quotient. Lift a primitive idempotent $\bar f$ of $\overline A$ to $f\in A$. The corner $fAf$ is a $kD$-linear direct summand of $A$ and is free over $kD$. Its quotient by $J(kD)$ is $\bar f\overline A\bar f\cong k$, so its rank is one. The central map
$$
kD\longrightarrow fAf,\qquad z\longmapsto zf
$$
is surjective by <Nakayama's lemma>, since $f$ generates its reduction, and is injective because both sides have dimension $|D|$. Thus $fAf\cong kD$ as algebras. Orthogonal primitive idempotents and their equivalences in the matrix quotient lift modulo the nilpotent ideal; lifting the equivalence maps and correcting their products by units lifts the matrix units. Consequently
$$
A\cong\operatorname{Mat}_r(fAf)\cong\operatorname{Mat}_r(kD).
$$
This proves the actual algebra isomorphism of <central defect blocks are matrix algebras>, as an <algebra>.
Finally suppose $D\triangleleft G$ but is not central. Put $C=C_G(D)$, which is normal in $G$ and contains $D$ since $D$ is abelian. The normal-subgroup argument in Question 5 puts the block idempotent $e_B$ in $kC$. It is therefore the sum of the $G$-conjugates of a block idempotent $c$ of $kC$. The extended first main theorem chooses such a $c$ with defect group $D$; since $D$ is normal, its pair stabilizer is exactly $I=\{g\in G:{}^gc=c\}$. Now
$$
e=[I:DC]=[I:C]=1,
$$
so $I=C$. Within $C$, the defect group $D$ is central, and the first case gives $kCc\cong\operatorname{Mat}_r(kD)$.
We can see the <Fong–Reynolds matrix decomposition> explicitly in this case. Choose representatives $g_1,\ldots,g_t$ for $G/C$, where $t=[G:C]$, and put
$$
u_{ij}=g_i c g_j^{-1}.
$$
Different conjugates of $c$ are orthogonal, so $u_{ij}u_{\ell m}=\delta_{j\ell}u_{im}$ and $\sum_i u_{ii}=e_B$. Moreover $cBc=ckGc=kCc$: a term $ckCgc$ vanishes unless $g$ stabilizes $c$, and that stabilizer is $C$. These matrix units therefore identify $B$ with $\operatorname{Mat}_t(kCc)$, whence
$$
\boxed{B\cong\operatorname{Mat}_{tr}(kD).}
$$
The inertial index condition concerns the stabilizer of the local block; it does not require every element of $G$ to centralize $D$. That distinction is why the second case is necessary.
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