(c)
First dissociation
$X(OH)_{3} \rightarrow X(OH)_{2}^{+} + OH^{-}$
Second dissociation
$X(OH)_{2}^{+} \rightarrow X(OH)^{2 +} + OH^{-}$
Total $\left\lbrack OH^{-} \right\rbrack = 4 \times 10^{- 3} + 2 \times 10^{- 3}$
$= 6 \times 10^{- 3}$
$pOH = 3 - \log 6$
$= 3 - 0.78 = 2.22$
$pH = 14 - 2.22 = 11.78$