$(1)$ $\lambda_{ m }^{\infty}( NaI )=12.7 \,m\,S m ^{2}\, mol ^{-1}$

$(2)$ $\lambda_{ m }^{\infty}\left( NaNO _{3}\right)=12.0 \,m\,S m ^{2}\, mol ^{-1}$

$(3)$ $\lambda_{ m }^{\infty}\left( AgNO _{3}\right)=13.3\,m\,S m ^{2}\, mol ^{-1}$

$\lambda_{ m }^{\infty}( Ag I )=(1)+(3)-(2)$

$=12.7+13.3-12.0$

$=26.0-12.0$

$\lambda_{ m }^{\infty}( Ag I )=14.0$