= Solution
Define $\gamma_{SV}$, $\gamma_{SL}$ and $\gamma_{LV}$ as the solid-vapor, solid-liquid and liquid-vapor interfacial free energies per actual area. For the corresponding planar material, <Young's contact-angle equation> is
$$
\gamma_{SV}-\gamma_{SL}=\gamma_{LV}\cos\theta.
$$
Consider advancing the macroscopic liquid footprint over projected area $A$, while retaining the specified wetting only to each sphere's equator. The newly wetted upper hemispheres have actual area $2\phi A$. The gaps occupy projected area $(1-\phi)A$, across which a liquid-vapor interface remains instead of solid-liquid contact.
The microscopic free-energy change, apart from the outer macroscopic droplet surface, is
$$
\frac{\Delta F}{A}
=2\phi(\gamma_{SL}-\gamma_{SV})
+(1-\phi)\gamma_{LV}.
$$
Equating this with $-\gamma_{LV}\cos\theta^*$, the macroscopic wetting contribution, yields
$$
\boxed{\cos\theta^*=2\phi\cos\theta-(1-\phi)
=\phi(2\cos\theta+1)-1.}
$$
The factor $2$ comes from the curved hemisphere area; the negative gap term is the trapped liquid-vapor interface. This combines Wenzel roughness on the wetted solid fraction with Cassie-type air gaps. It differs from both homogeneous Wenzel wetting and a Cassie formula that treats the solid contacts as flat.
For $\phi=0.9069$ and $\theta=68^\circ$, this gives $\cos\theta^*\simeq0.5864$ and $\theta^*\simeq54.1^\circ$. The hybrid state is less wettable than the fully penetrated hemispherical model, but still not hydrophobic. The result is conditional on the assumed equatorial meniscus and retained air: pinning or metastability may be required, and allowing further penetration can change the equilibrium geometry.
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