= Solution
Use <exposure dose> $D(x)=I(x)\tau$, with $\mathrm{mW}\times\mathrm s=\mathrm{mJ}$. At approximately $x=1.0$ and $3.0\,\mu\mathrm m$, the positive resist just starts to thin. The intensity there is about $5\,\mathrm{mW\,cm^{-2}}$, so
$$
\boxed{\tau=\frac{D_0}{I}=\frac{10\,\mathrm{mJ\,cm^{-2}}}{5\,\mathrm{mW\,cm^{-2}}}=2.0\,\mathrm s.}
$$
As a consistency check, the fully cleared region begins at approximately $x=1.5$ and ends at $2.5\,\mu\mathrm m$, where the intensity is about $25\,\mathrm{mW\,cm^{-2}}$. Multiplying by $2\,\mathrm s$ gives the clearing dose $50\,\mathrm{mJ\,cm^{-2}}$ at both edges.
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