If the ionic product of $Ni(OH{)}_2$ is $1.9\times {10}^{–15}$, the molar solubility of $Ni(OH{)}_2$ in 1.0 M NaOH is

$Ni(OH{)}_2\rightleftharpoons N{i}^{2+}+O{H}^{-}$

$\left[N{i}^{2+}\right]$ = s, $\left[O{H}^{-}\right]$ = 2s

$NaoH\rightleftharpoons N{a}^{+}+O{H}^{-}$

$\left[N{a}^{+}\right]$= 1, $\left[O{H}^{-}\right]$ = 1

Total $\left[O{H}^{-}\right]$ = 2s + 1

$K_{s}p=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$

= s (2s + 1)

As $K_{sp}$  is small, 2s<<1

$K_{s}p=s\times 1^2$

$1.9\times {10}^{–15}$ = s

Hence, the correct option is option (C) $1.9\times {10}^{-15}$ M.