At 298 K the cell Pt | H2(1 bar) | HA(aq, 0.10 M) || Ag^+(aq, 0.10 M) | Ag has an emf of 0.918 V. Take E^°Ag^+/Ag=0.80 V and 2.303RT/F=0.059 V, and…

The hydrogen electrode is the anode and the silver electrode the cathode, so $E_{cell}=E_{\mathrm{Ag^+/Ag}}-E_{\mathrm{H^+/H_2}}$. For silver, $E=0.80+0.059\log(0.10)=0.741\ \mathrm{V}$. For the hydrogen electrode at 1 bar, $E=-0.059\ \mathrm{pH}$. Hence $0.918=0.741+0.059\,\mathrm{pH}$, giving $\mathrm{pH}=3.00$ and $[\mathrm{H^+}]=1.0\times10^{-3}\ \mathrm{M}$. For $\mathrm{HA\rightleftharpoons H^++A^-}$, $K_a=[\mathrm{H^+}]^2/(0.10-[\mathrm{H^+}])=10^{-6}/0.099\approx1.0\times10^{-5}$. The wrong choices come from forgetting the Nernst correction for $\mathrm{Ag^+}$ (which uses 0.80 V and gives pH 2, $K_a\approx1.1\times10^{-3}$), or from dropping the square in $[\mathrm{H^+}]^2$ or dividing by the wrong concentration. Remember also that the hydrogen electrode potential is negative of $0.059\,\mathrm{pH}$, so the cell emf rises with pH, and each pH unit contributes $0.059\ \mathrm{V}$. Only option (C), $1.0\times10^{-5}$, follows from the full treatment of both half-cells and the weak-acid equilibrium.