We know,
$\mathrm{pH} = -\log \left[{\mathrm{H}}^{+}\right]$ , $\mathrm{pH} = -\log \left[{\mathrm{OH}}^{-}\right], \mathrm{pH} + \mathrm{pOH} = 14$
$\because \mathrm{pH}=12$

$\therefore \left[{\mathrm{H}}^{+}\right]={10}^{-12}\mathrm{M}$

$\therefore \left[{\mathrm{OH}}^{-}\right]={10}^{-2}\mathrm{M}$
We also know,
$\mathrm{Ca}(\mathrm{OH}{)}_2 \rightleftharpoons 2{\mathrm{OH}}^{-} + {\mathrm{Ca}}^{2+}$

$\therefore \left[\mathrm{Ca}(\mathrm{OH}{)}_2\right]=5\times {10}^{-3}\mathrm{M}$

$5\times {10}^{-3}=\frac{\text{milli moles of} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 }{100\mathrm{mL}}$

milli moles of $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2=5\times {10}^{-1}$

Ans. $=5$