The pH of 0.1M weak mono acidic base is 9. The dissociation constant of the base is

 pH + pOH = 14

9 + pOH = 14

pOH = 14 – 9 = 5

pOH = $-log[O{H}^{[}-\}]$

5 = $-log[O{H}^{[}-\}]$

[$O{H}^{-}]={10}^{-5}$

Consider a weak mono acidic base,

$BOH\rightleftharpoons B^{+}+O{H}^{-}$

$K_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{BOH}$

Since, 0.1 >> ${10}^{-5}$, [BOH]=0.1 mol L^ {−1}

& Also $\left[B^{+}\right]=\left[O{H}^{-}\right]$

Therefore, $K_b=\frac{[O{H}^{-}{]}^2}{BOH}$

= $\frac{({10}^{-5}{)}^2}{0.1}$

= $1\times {10}^{-9}$

Hence, the correct option is option (B) ${10}^{–9}$.