= Solution
At the <photosphere>, use the stated boundary pressure and the <effective temperature> relation $L=4\pi R^2\sigma T_{\rm eff}^4$. The <radiation constant> obeys $a_{\rm rad}c=4\sigma$ by the <Stefan–Boltzmann law>. Substitution into the <radiative temperature gradient> gives
$$
\nabla_{\rm phot}=\frac{3\kappa_{\rm phot}L}{16\pi a_{\rm rad}cGM T_{\rm eff}^4}\frac{2GM}{3\kappa_{\rm phot}R^2}
=\frac{2\sigma}{4a_{\rm rad}c}=\boxed{\frac18}.
$$
This uses the approximate <stellar surface boundary condition> at optical depth $2/3$; it is not a universal photospheric gradient for arbitrary atmospheres.
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