= Solution
Part (a), applied to any <ideal class>, produces an <integral ideal> of positive <integer> <norm> at most $c_K$, so $c_K\ge1$. Writing $n=[K:\mathbb Q]$ and using $s\le n/2$ gives
$$
\sqrt{|\Delta^2(K)|}\ge\left(\frac\pi4\right)^s\frac{n^n}{n!}
\ge\left(\frac\pi4\right)^{n/2}2^{\lfloor n/2\rfloor}.
$$
For the last inequality, write $n^n/n!=\prod_{j=1}^n(n/j)$; the first $\lfloor n/2\rfloor$ factors are at least $2$, and the remaining factors are at least $1$. Since $\lfloor n/2\rfloor\ge(n-1)/2$, squaring yields
$$
|\Delta^2(K)|\ge\frac12\left(\frac\pi2\right)^n.
$$
As $\pi/2>1$, the assumed upper bound $D$ forces
$$
\boxed{n\le\frac{\log(2D)}{\log(\pi/2)}.}
$$
For $D\ge1$, the ceiling of the right-hand side is a valid <integer> $M$. If $D=0$, no number <field> satisfies the premise, so any positive $M$ works.
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