For AgCl,
\(\begin{array}{*{20}{c}}{{K_{sp}}(AgCl) = \left( {A{g^ + }} \right)\left( {C{l^ - }} \right) = 1.8 \times {{10}^{ - 10}}}\\{\left( {A{g^ + }} \right) = \frac{{1.8 \times {{10}^{ - 10}}}}{{0.10M}} = 1.8 \times {{10}^{ - 9}}M}\end{array}\)
For AgI,
\(\begin{array}{*{20}{c}}{{K_{sp}}(AgI) = \left( {A{g^ + }} \right)\left( {{I^ - }} \right) = 1.5 \times {{10}^{ - 16}}}\\{\left( {A{g^ + }} \right) = \frac{{1.5 \times {{10}^{ - 16}}}}{{1 \times {{10}^{ - 2}}M}} = 1.5 \times {{10}^{ - 9}}M}\end{array}\)
\(\left( {A{g^ + }} \right)(AgCl) > \left( {A{g^ + }} \right)(AgI)\)
Therefore the solution is solid AgI will form first.