= Solution
Put $\theta(r)=\operatorname{Vol}B(p,r)/(\omega_nr^n)$. The <Bishop-Gromov inequality> says that $\theta$ is nonincreasing and at most one, while smoothness gives $\theta(r)\to1$ as $r\downarrow0$. The prescribed limit at infinity is also one. Hence $\theta(r)=1$ for every $r>0$.
Here is how equality gives the global <Euclidean rigidity of maximal asymptotic volume ratio>. Assume first $n\geq2$. In <geodesic polar coordinates> centered at $p$, let $c(\xi)$ be the cut time along unit direction $\xi$, and $J(r,\xi)$ the positive radial volume density before that time. The comparison giving the <Bishop-Gromov inequality> says $J(r,\xi)\leq r^{n-1}$, and
$$
\operatorname{Vol}B(p,R)=\int_{S^{n-1}}\int_0^{\min\{R,c(\xi)\}}J(r,\xi)\,dr\,d\xi.
$$
No cut time can be finite. If $c(\xi_0)<t$, then the radial segment at time $t$ is not minimizing, so $d(p,\exp_p(t\xi_0))<t$. This strict inequality persists for nearby directions by continuity, and their cut times are also at most $t$. At any radius $R>t$, that open set of directions omits all radial volume from $t$ to $R$. Even the maximal density $r^{n-1}$ cannot then give Euclidean volume, contradicting $\theta(R)=1$. Thus $c(\xi)=\infty$ for all directions. Equality of all ball volumes and continuity now force $J(r,\xi)=r^{n-1}$ everywhere.
Let $A(r)X=\nabla_X\partial_r$ on the distance sphere. This is the negative of the outward <shape operator> in the convention $S=-\nabla\nu$. Put $h=\operatorname{tr}A=\partial_r\log J$. The traced <radial Riccati equation for distance spheres> is
$$
h'+\operatorname{tr}(A^2)+\operatorname{Ric}_{\mathrm{tr}}(\partial_r,\partial_r)=0.
$$
Since $h=(n-1)/r$, it gives
$$
\operatorname{tr}(A^2)+\operatorname{Ric}_{\mathrm{tr}}(\partial_r,\partial_r)=\frac{n-1}{r^2}.
$$
But $\operatorname{tr}(A^2)\geq h^2/(n-1)=(n-1)/r^2$ by the <Cauchy-Schwarz inequality>, and the Ricci term is nonnegative. Equality therefore holds in both, so $A=r^{-1}I$. Writing the polar metric as $dr^2+g_r$, its evolution is $\partial_rg_r=2g_r/r$. Its small-radius Euclidean limit then gives
$$
g_r=r^2g_{S^{n-1}},\qquad g=dr^2+r^2g_{S^{n-1}}.
$$
The <Riemannian exponential map> is onto by the <Hopf-Rinow theorem>, has no conjugate points, and is injective because there is no finite cut time. Explicitly, two distinct minimizing radial <geodesics> reaching the same point would, after extending one past that point, create a length-minimizing broken curve with a genuine corner; smoothing the corner shortens it. That contradicts global minimization of every radial segment. Thus $\exp_p$ is a global <diffeomorphism>, and the displayed metric proves
$$
\boxed{(M,g)\cong(\mathbb R^n,g_{\mathrm{Euclidean}})\text{ isometrically}.}
$$
In dimension one, a <connected> complete manifold without boundary is a line or a circle. The circle has limiting volume ratio zero, so the hypothesis leaves only the Euclidean line.
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