Explanation

(b)
The compound which is having least solubility will be precipitated first.
$BaSO_{4}$ : Given, $K_{sp} = 10^{- 11}$
Let the solubility=$x$ mol/L
$BaSO_{4} \rightarrow Ba^{2}\ + SO_{4}^{2 -}\$
$x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x$
$\therefore$ $K_{sp} = \left\lbrack Ba^{2 +} \right\rbrack\lbrack SO_{4}^{2 -}\rbrack$
or $K_{sp} = x \times x$
$\therefore$ $K_{sp} = x^{2}$
or $x = \sqrt{K_{sp}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{10^{- 11}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.16 \times 10^{- 6}$ mol/L
$CaSO_{3}$ : Given, $K_{sp} = 10^{- 6}$
Let the solubility=$x$ mol/L
$CaSO_{4} \rightarrow \lbrack Ca^{2 +}\rbrack\lbrack SO_{4}^{2 -}\rbrack$
$\therefore$ $K_{sp} = \left\lbrack Ca^{2 +} \right\rbrack\lbrack So_{4}^{2 -}\rbrack$
or $K_{sp} = x \times x$
$\therefore$ $K_{sp} = x^{2}$
or $x = \sqrt{K_{sp}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{10^{- 6}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 10^{- 3}$ mol/L
$AgSO_{4}$ : Given, $K_{sp} = 10^{- 5}$
Let the solubility=$x$ mol/L
$Ag_{2}SO_{4} \rightarrow 2Ag^{+} + SO_{4}^{2 -}$
$2x$ $x$
$\therefore$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{sp} = \left\lbrack Ag^{+} \right\rbrack^{2}\lbrack SO_{4}^{2 -}\rbrack$
or $= (2x)^{2}(x)$
or $K_{sp} = 4x^{3}$
or $x = \frac{\sqrt[3]{K_{sp}}}{4}$
$= \frac{\sqrt{10^{- 5}}}{4}$
$= 10^{- 2}$ mol/L
$\because$ $BaSO_{4}$ has least solubility.
$\ \therefore$ It will precipitated first.