= Solution
For a fixed center height $h$, the radius decreases according to the preceding current enhancement. Thus
$$
\tau_h(R_0)=\frac{2\phi_B^2}{M\gamma}\int_0^{R_0}\frac{R^2}{c(R/h)}\,dR.
$$
Within the stipulated logarithmic approximation, $c(y)=1+y/2+O(y^2)$, so $1/c(y)=1-y/2+O(y^2)$. Expanding uniformly over $0\leq R\leq R_0\ll h$ and integrating gives
$$
\tau_h=\frac{2\phi_B^2}{M\gamma}\left[\frac{R_0^3}{3}-\frac{R_0^4}{8h}+O\left(\frac{R_0^5}{h^2}\right)\right].
$$
Therefore the leading correction to the isolated-droplet lifetime is
$$
\boxed{\tau_h(R_0)=\tau_\infty(R_0)\left[1-\frac{3R_0}{8h}+O((R_0/h)^2)\right],\qquad
\Delta\tau=-\frac{\phi_B^2R_0^4}{4M\gamma h}.}
$$
The negative correction reflects faster evaporation into the nearby reservoir. This leading coefficient also agrees with the small-$R/h$ expansion of the exact <sphere-plane capacitance>; the stipulated logarithmic approximation need not match higher orders.
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