The solubility of AgSCN in 0.002 M $N{H}_3$ is ($K_{sp}$ for $AgSCN=1.0\times {10}^{-12};K_d$ for $Ag{\left(N{H}_3\right)}_2^{+}=1.0\times {10}^{-8}$ )

$Ag{\left(N{H}_3\right)}_2^{+}\stackrel{}{\rightleftharpoons }A{g}^{+}+2N{H}_3,$

$K_d=1.0\times {10}^{-8}=\frac{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}{\left[Ag{\left(N{H}_3\right)}_2^{+}\right]}$

If $smol{L}^{-1}=$ solubility of $AgSCN,\left[Ag{\left(N{H}_3\right)}_2^{+}\right]$

$=S\left[SC{N}^{-}\right]=S$

Hence, $=\frac{\left[A{g}^{+}\right]{\left[0.002\right]}^2}{S}=1.0\times {10}^{-8}$

$\Rightarrow \left[A{g}^{+}\right]=\frac{1.0\times {10}^{-8}S}{4\times {10}^{-6}}=2.5\times {10}^{-3}S$

Also, $K_{sp}\left[AgSCN\right]=1.0\times {10}^{-12}=\left[A{g}^{+}\right]\left[SC{N}^{-}\right]$

$=2.5\times {10}^{-3}S\times S\Rightarrow S=2.0\times {10}^{-5}mol{L}^{-1}$