= Solution
First derive the <Laplace functional of a Poisson random measure>. For a simple nonnegative $g=\sum_jt_j\mathbf1_{A_j}$ on disjoint finite-intensity sets, independent <Poisson random variables> give
$$
\mathbb E e^{-M(g)}=\prod_j\exp\bigl(\mu(A_j)(e^{-t_j}-1)\bigr)
=\exp\left(-\int(1-e^{-g})\,d\mu\right).
$$
Increasing simple approximations, <monotone convergence> for the integrals, and <bounded convergence theorem> for the random exponentials prove the same identity for every nonnegative measurable $g$.
Here the intensity is <Lebesgue measure>. Put $A=B(0,r)$ and
$$
L=\exp\left(-\int_{\mathbb R^d}(1-e^{-f(x)})\,dx\right)=\mathbb E e^{-M(f)}.
$$
The integral is finite because $f$ has compact support. Apply the <Laplace functional of a Poisson random measure> to $f+t\mathbf1_A$, $t\geq0$:
$$
\mathbb E e^{-M(f)-tN_r}
=L\exp\left(-(1-e^{-t})\int_Ae^{-f(x)}\,dx\right).
$$
Differentiate from the right at $t=0$. The difference quotient on the random side is dominated by $N_r$, which is integrable with mean $v_dr^d$; the intensity integral is over a finite-volume ball. We obtain the <count-weighted Laplace functional of a Poisson random measure>:
$$
\boxed{\mathbb E[N_re^{-M(f)}]
=\exp\left(-\int_{\mathbb R^d}(1-e^{-f(x)})\,dx\right)\int_{B(0,r)}e^{-f(x)}\,dx.}
$$
At $r=0$, the ball is empty and both sides are zero.
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