b
\(\mathrm{Ag}_{2} \mathrm{CrO}_{4} \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_{4}^{2-}\)

solubility product \(\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{s})^{2} \times \mathrm{S}=4 \mathrm{s}^{3}\)

\(\mathrm{K}_{\mathrm{sp}}=\left(1.1 \times 10^{-12}\right)\)

\(s=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=0.65 \times 10^{-4}\)

\(\mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}\)

\(\mathrm{K}_{\mathrm{sp}}=5 \times 5 \quad\left(\mathrm{K}_{\mathrm{sp}}=1.8 \times 10^{-10}\right)\)

\(s=\sqrt{\mathrm{K}_{\mathrm{sp}}}=1.34 \times 10^{-5}\)

AgBr \(\rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\)

\(\mathrm{K}_{\mathrm{sp}}=5 \times 5 \quad\left(\mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13}\right.\)

\(S=\sqrt{K_{s p}}=0.71 \times 10^{-6}\)

\(\mathrm{Agl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-}\)

\(\left.\mathrm{Ksp}=\mathrm{S} \times \mathrm{S} \quad \mathrm{K}_{\mathrm{sp}}=8.3 \times 10^{-17}\right)\)

\(S=\sqrt{K_{\mathrm{sp}}}=0.9 \times 10^{-8}\)

therefore, solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is highest so it will precipitate last.