= Solution
The depth-averaged added horizontal compressive <stress> is $\bar\sigma=F/a$. Resolving its <traction> on a plane dipping at angle $\delta$ gives the <fault shear stress from a plate-integrated force>,
$$
\bar\tau=\bar\sigma\sin\delta\cos\delta,
\qquad
\boxed{|\bar\tau|=\frac{F}{2a}\simeq1.30\times10^7\,\mathrm{Pa}=13\,\mathrm{MPa}
\quad(\delta=45^\circ).}
$$
Keeping the previous topographic correction gives about $13.8\,\mathrm{MPa}$. This is the mean additional shear from <ridge push>, assuming the <force> is transmitted through thickness $a$ and resolving the nonisotropic <stress> increment. An isotropic lithostatic <pressure> adds normal <traction> but no shear. The <force> <integral> alone does not specify the local <stress> at each point on the <fault> or establish failure.
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