= Solution
Choose the bounded, nonnegative <nondecreasing function> $\tau$ from part (d), and define
$$
h(r)=\int_0^r\tau(s)\,ds,\qquad U(x)=h(|x|).
$$
The <function> $h$ is <convex> and nondecreasing. For $0\leq a\leq1$, the <triangle inequality>, monotonicity, and <convexity> give
$$
U((1-a)x+ay)\leq h((1-a)|x|+a|y|)\leq(1-a)U(x)+aU(y).
$$
Hence $U$ is a <convex function>. By the <gradient of a radial function> formula and the <fundamental theorem of calculus>,
$$
\nabla U(x)=\tau(|x|)\frac{x}{|x|}
$$
for $x\ne0$ whenever $h'(|x|)=\tau(|x|)$. A <monotone function> has at most countably many discontinuities, so part (d) makes the exceptional radii $\widetilde\mu$-null. The origin is also $\mu$-null. Define $\mathcal T(0)=0$; thus $\mathcal T=\nabla U$ $\mu$-almost everywhere.
For every $Q$ in the <orthogonal group>, $\mathcal T(Qx)=Q\mathcal T(x)$, including at zero. Consequently $\mathcal T_\#\mu$ is invariant under the <orthogonal group>. Moreover $|\mathcal T(x)|=\tau(|x|)$ away from the null origin, and composition of <pushforward measures> gives
$$
\Phi_\#(\mathcal T_\#\mu)=\tau_\#(\Phi_\#\mu)=\widetilde\nu=\Phi_\#\nu.
$$
Part (c) therefore gives $\mathcal T_\#\mu=\nu$. The <Knott–Smith optimality criterion> makes a <transport map> given by a <convex function>'s <gradient> optimal for the quadratic <transport cost>. Uniqueness from <Brenier's theorem> yields
$$
\boxed{T(x)=\mathcal T(x)=\tau(|x|)\frac{x}{|x|}\quad\mu\text{-almost everywhere}.}
$$
Since directions agree, this also identifies the optimal value:
$$
\boxed{d_2^2(\mu,\nu)=\int_0^\infty|r-\tau(r)|^2\,d\widetilde\mu(r)=d_2^2(\widetilde\mu,\widetilde\nu).}
$$
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