Solubility of $\text{AgBr}$ in $\text{0}\text{.01}\text{M}{\text{CaBr}}_{\text{2}}\left({\text{S}}_{\text{2}}\right)$ ......(i)

${\text{CaBr}}_{\text{2}}$ is a strong electrolyte and dissociates completely to give common ion ${\text{Br}}^{-}$

$\underset{0.01\text{M}}{{\text{CaBr}}_{\text{2}}}\to \underset{\text{0}\text{.01}\text{M}}{{\text{Ca}}^{\text{2+}}}+\underset{\text{0}\text{.02}\text{M}}{{\text{2Br}}^{-}}$

$\text{AgBr(s)}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{(aq)+Br}}^{-}\text{(aq);}{\text{K}}_{\text{sp}}$

${\text{S}}_{\text{2}}\text{0}{\text{.02+S}}_{\text{2}}\Rightarrow {\text{S}}_{\text{2}}\times 0.02={\text{K}}_{\text{sp}}$

$~0.02$ or ${\text{S}}_{\text{2}}=50{\text{K}}_{\text{sp}}$

Similarly, $\text{KBr}$ gives common ion ${\text{Br}}^{-}$

And ${\text{AgNO}}_{\text{3}}$ gives common ion ${\text{Ag}}^{\text{+}}$ and we can calculate

${\text{S}}_{\text{3}}\text{=}{\text{100 K}}_{\text{sp}}$ ......(iii)

And ${\text{S}}_{\text{4}}=20{\text{K}}_{\text{sp}}$ ......(iv) respectively

From (i), (ii), (iii) and (iv)

${\text{S}}_{\text{1}}\text{>}{\text{S}}_{\text{3}}\text{>}{\text{S}}_{\text{2}}\text{>}{\text{S}}_{\text{4}}$

(Note: $\sqrt{{\text{K}}_{\text{sp}}}\text{>}\text{100}{\text{K}}_{\text{sp}}\because {\text{K}}_{\text{sp}}$ order $-{10}^{-12}$ )