= Solution
{c}
Take $D=\langle(12)(34)\rangle$ and
$$
P=\{1,(12)(34),(13)(24),(14)(23)\}.
$$
Every involution in $A_5$ is a double transposition. They form one conjugacy class: an odd $S_5$ conjugator can be made even by multiplying by a transposition centralizing the source double transposition. Thus all order-two subgroups of $A_5$ are conjugate. Since $|A_5|=60$, $P$ is a <Sylow subgroup>. A permutation normalizing $P$ fixes its unique common fixed letter, $5$, and every even permutation of the other four letters permutes its three double transpositions. Hence $N_{A_5}(P)=A_4$. Any permutation centralizing $(12)(34)$ fixes $5$; within $S_4$ it may exchange the two pairs and interchange letters in either pair, giving eight possibilities, of which exactly the four elements of $P$ are even. Thus $C_{A_5}(D)=P$. Since $P$ is abelian and contains $D$, also $C_{A_5}(P)=P$.
The given single block of $kA_4$ has identity $1$ and defect $P$, since $\operatorname{Br}_P^{A_4}(1)=1$ and $P$ is Sylow. By <Brauer first main theorem>, $kA_5$ has exactly one block with defect $P$.
Suppose a block idempotent $b$ had defect $D$. Then $\operatorname{Br}_D^{A_5}(b)$ is a nonzero idempotent in $kP$. A <group algebra of a p-group in characteristic p is local>, so this image is $1$. But both the Brauer projections at $D$ and at $P$ retain exactly the basis elements of the same centralizer $P$. Consequently
$$
\operatorname{Br}_P^{A_5}(b)=\operatorname{Br}_D^{A_5}(b)=1.
$$
Thus $b$ has the full Sylow defect $P$, contradicting maximality of $D$. These are the <2-modular defect groups of A5>, and $\boxed{C_2\text{ cannot be a defect group}}$.
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