 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 $5\mathrm{mW}\mathrm{c}{\mathrm{m}}^{-2}$, soAs 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 $25\mathrm{mW}\mathrm{c}{\mathrm{m}}^{-2}$. Multiplying by $2\mathrm{s}$ gives the clearing dose $50\mathrm{mJ}\mathrm{c}{\mathrm{m}}^{-2}$ at both edges.