= Solution
In a sufficiently <uniform flow> the <Stokes law>[Stokes force] is $F=\gamma v$ toward trap 2. Its <work> over separation $d$ lowers the second effective well by $Fd$. Equal effective free energies require $\gamma vd=k_BT\log2$. Hence
$$
\boxed{v=\frac{k_BT\log2}{6\pi\eta a d}\simeq3.0\times10^{-7}\ \mathrm{m\,s^{-1}}}
$$
at 298 K. The <temperature> was not specified, so the numerical result scales as $T/298\ \mathrm K$. This approximation treats a weak uniform <drag (physics)>[drag] as a constant tilt and neglects changes in barrier shape and <Hydrodynamic interaction>[hydrodynamic interactions] with boundaries.
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