= Solution
For $r\geq0$, <Lebesgue measure> of the ball is $v_dr^d$, so the <Poisson random measure> gives $N_r\sim\operatorname{Poisson}(v_dr^d)$ and
$$
\mathbb P(N_r=0)=e^{-v_dr^d}.
$$
By monotonicity of the balls, the set of radii with no points is an interval starting at zero. Thus $\{R\geq r\}=\{N_r=0\}$: if all smaller balls are empty then their union, the open ball of radius $r$, is empty; if that ball is empty then its radius belongs to the defining set. We obtain
$$
\mathbb P(R\geq r)=e^{-v_dr^d}.
$$
The continuous tail gives $\mathbb P(R=0)=0$ and $\mathbb P(R=\infty)=0$. Differentiating the <cumulative distribution function> gives the <probability density function>
$$
\boxed{f_R(r)=d v_d r^{d-1}e^{-v_dr^d}\mathbf1_{\{r>0\}}.}
$$
The substitution $u=v_dr^d$ proves that this density integrates to one; equivalently $v_dR^d$ has the unit-rate <exponential distribution>. This is the first-neighbour case of the <Kth-nearest-neighbour distance in a homogeneous Poisson point process>.
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