For precipitation,
ionic product $>$ solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$
For ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$
$$\begin{array}{rl}\text{ ionic product }& ={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CrO}}_4^{-}\right]\\ & ={\left({10}^{-4}\right)}^2\left({10}^{-5}\right)={10}^{-13}\end{array}$$
${\mathrm{K}}_{\mathrm{sp}}$ of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4=4\times {10}^{-12}$
Here, ${\mathrm{K}}_{\mathrm{sp}}>\mathrm{IP}$
Thus, no precipitate is obtained.
For $\mathrm{AgCl}$,
ionic product $=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]$
$=\left[{10}^{-4}\right]\left[{10}^{-5}\right]$
$={10}^{-9}$
${\mathrm{K}}_{\mathrm{sp}}(\mathrm{AgCl})=1\times {10}^{-10}$
Here, IP > ${\mathrm{K}}_{\mathrm{sp}}$
So, precipitate will form.
Thus, silver chloride gets precipitated first.