$n_{Mg^{2+}} = 0,003$ mol ; $n_{Ca^{2+}} = 0,002$ mol
$Mg(HCO_3)_2 + Ca(OH)_2 => Mg(OH)_2 + Ca(HCO_3)_2$$0,003 0,003 0,003$
$Ca(HCO_3)_2 + Ca(OH)_2 => 2CaCO_3 + 2H_2O$
$0,005 0,005$
=> $n_{Ca(OH)_2} = 0,008$ mol => $V = 0,008/0,05 = 0,16$ lít = $160$ml