  Let $B_s$ be a three dimensional Brownian motion and $\omega(dx)$ be an
independent Poisson field on $\mathbb{R}^3$. It is proved that for any $t>0$,
conditionally on $\omega(\cdot)$,
  \label{*} \mathbb{E}_0 \exp\{\theta \int_0^t \bar{V}(B_s) ds\} \ < \infty \
a.s. & \text{if} \theta< 1/16, \medskip
  = \infty \ a.s. & \text{if} \theta> 1/16, where $\bar{V}(x)$ is the
renormalized Poisson potential
  $$ \bar{V}(x)=\int_{\mathbb{R}^3} \frac{1}{| x-y |^2}
\big[\omega(dy)-dy\big]. $$ Then the long term behavior of the quenched
exponential moment \eqref{*} is determined for $\theta \in (0, 1/16)$ in the
form of integral tests.
  This paper exhibits and builds upon the interrelation between the exponential
moment \eqref{*} and the celebrated Hardy's inequality $$ \int_{\mathbb{R}^3}
\frac{f^2(x)}{| x |^2} dx \le 4 \|\nabla f\|_2^2, 2in f \in
W^{1,2}(\mathbb{R}^3). $$
