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An optimal partition $Q=(B_1,B_2,B_3)$ is as follows:
$B_j=\big(\sqrt\frac j3\,D\big)\setminus\big(\sqrt\frac{j-1}3\, D\big)$ for $j=1,2,3$. Moreover, the optimal partition is unique, up to sets of zero Lebesgue measure.
To begin proving these statements, note that the $B_j$'s are the annuli of area $1$ each centered at the origin, with the following property:
whenever $1\le j_1<j_2\le 3$ and $E_1$ and $E_2$ are Borel sets of nonzero Lebesgue measure such that $E_1\subseteq B_{j_1}$ and $E_2\subseteq B_{j_2}$, we have
\begin{equation*}
\ave_{E_1}f\ge\ave_{E_2}f, \tag{0}
\end{equation*}
where $f$ is the density of the bivariate standard normal probability distribution $\mu$ with respect to the Lebesgue measure and
\begin{equation*}
\ave_E f:=\frac{\mu(E)}{|E|}=\frac1{|E|}\int_E f(x)dx,
\end{equation*}
the average value of density $f$ over the set $E$.
Property (0) holds because $\inf_{B_{j_1}}f\ge \sup_{B_{j_2}} f$ if $1\le j_1<j_2\le 3$.
Take now any partition $P=(A_1,A_2,A_3)\in\mathcal P$ and let $C_{i,j}:=A_i\cap B_j$. The crucial observation is that the matrix $(|C_{i,j}|)_{i,j=1}^3$ is double stochastic, and so, by the Birkhoff--von Neumann theorem,
\begin{equation*}
|C_{i,j}|=\sum_{\pi\in\Pi}w_\pi \ii{j=\pi(i)}
\end{equation*}
for all $i,j$,
where $\Pi$ is the set of all permutations of the set $\{1,2,3\}$, $\ii{\cdot}$ is the indicator function, and the $w_\pi$'s are some nonnegative real numbers such that $\sum_{\pi\in\Pi}w_\pi=1$.
Therefore and because the Lebesgue measure has no atoms, for each triple $(i,j,\pi)$ there is a Borel set $C_{\pi;i,j}$ such that
\begin{equation*}
|C_{\pi;i,j}|=w_\pi \ii{j=\pi(i)} \tag{1}
\end{equation*}
and $(C_{\pi;i,j})_{\pi\in\Pi}$ is a partition of $C_{i,j}$ for each $(i,j)$.
Let now
\begin{equation*}
r_{\pi;i,j}:=\frac{\mu(C_{\pi;i,j})}{|C_{\pi;i,j}|}=\ave_{C_{\pi;i,j}} f
\end{equation*}
if $|C_{\pi;i,j}|\ne0$; otherwise, define $r_{\pi;i,j}$ arbitrarily, but so that one have
\begin{equation}
r_{\pi;i_1,j_1}\ge r_{\pi;i_2,j_2} \tag{2}
\end{equation}
whenever $1\le j_1<j_2\le 3$, for any $\pi,i_1,i_2$; this is possible to do because of the property highlighted above; for instance, one may let $r_{\pi;i,j}:=\sup_{B_j}f$ for all $\pi,i,j$ such that $|C_{\pi;i,j}|=0$.
Then
\begin{multline}
S(P)=\sum_{i=1}^3\,i\,\mu(A_i)
=\sum_{i,j=1}^3\,i\,\mu(C_{i,j})
=\sum_\pi\sum_{i,j}\,i\,\mu(C_{\pi;i,j})
=\sum_\pi\sum_{i,j}\,i\,r_{\pi;i,j}|C_{\pi;i,j}| \\
=\sum_\pi w_\pi\sum_i\,i\,r_{\pi;i,\pi(i)}
=\sum_\pi w_\pi\sum_j\,\pi^{-1}(j)\,a_{\pi;j} \tag{3}
\end{multline}
by (1), where
\begin{equation*}
a_{\pi;j}:=r_{\pi;\pi^{-1}(j),j}. \tag{4}
\end{equation*}
Similarly, with $Q=(B_1,B_2,B_3)$ as before,
\begin{multline}
S(Q)=\sum_{i=1}^3\,j\,\mu(B_j)
=\sum_{i,j=1}^3\,j\,\mu(C_{i,j})
=\sum_\pi\sum_{i,j}\,j\,\mu(C_{\pi;i,j})
=\sum_\pi\sum_{i,j}\,j\,r_{\pi;i,j}|C_{\pi;i,j}| \\
=\sum_\pi w_\pi\sum_j\,j\,r_{\pi;\pi^{-1}(j),j}
=\sum_\pi w_\pi\sum_j\,j\,a_{\pi;j}. \tag{5}
\end{multline}
Fix now any $\pi\in\Pi$ and let $\si:=\pi^{-1}$, $a_j:=a_{\pi;j}$, and $h_i:=a_i-a_{i+1}$ (with $a_4:=0$).
By (4) and (2), $h_i\ge0$ for all $i=1,2,3$. So,
\begin{multline}
\sum_j\,\pi^{-1}(j)\,a_{\pi;j}=\sum_j\,\si(j)\,a_j=\sum_{j=1}^3\,\si(j)\,\sum_{i=j}^3 h_i
=\sum_{i=1}^3\,h_i\sum_{j=1}^i\si(j) \\
\ge\sum_{i=1}^3\,h_i\sum_{j=1}^i j
=\sum_{j=1}^3\,j\,\sum_{i=j}^3 h_i
=\sum_j\,j\,a_j=\sum_j\,j\,a_{\pi;j}.
\end{multline}
So, by (3) and (5), $S(P)\ge S(Q)$, for any $P\in\mathcal P$. That is, the partition $Q$ is optimal. Following the lines of the above proof, we can see that the optimal partition is unique, up to sets of zero Lebesgue measure. This concludes the entire proof.
