= Solution
Let $A$ be projected area. Close-packed circular footprints cover $\phi A$ and the exposed flat substrate covers $(1-\phi)A$. A hemisphere's curved area is twice its circular footprint, so the actual wetted solid area is
$$
A_{\mathrm{wet}}=2\phi A+(1-\phi)A=(1+\phi)A.
$$
The <Wenzel roughness factor> is therefore $r=A_{\mathrm{wet}}/A=1+\phi=1.9069$, not $\phi$ or $2$.
Using the <Wenzel wetting> relation,
$$
\cos\theta^*=1.9069\cos68^\circ\simeq0.7143,
\qquad \theta^*\simeq44.4^\circ.
$$
The nanostructure makes this surface \b[more wettable], because planar PMMA already has $\theta<90^\circ$. Under complete Wenzel wetting, roughening amplifies the underlying hydrophilic character; it does not turn this material hydrophobic.
The calculation assumes identical chemistry on spheres and substrate, complete liquid penetration, ideal hemispheres, a macroscopic contact angle and negligible pinning/line-tension effects. Actual dried particles may have different contact geometry, deformation, contamination or trapped air. A hydrophobic coating with planar angle above $90^\circ$ would make roughness increase the angle instead. Designing a texture that retains air in a <Cassie–Baxter state> is another route, but its stability against liquid penetration must be checked.
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