$CaO+{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}$

0,3             $\to $ 0,3                (mol)

${{n}_{CaO}}=\frac{{{m}_{CaO}}}{{{M}_{CaO}}}=\frac{16,8}{56}=0,3\left( mol \right)$

Theo phương trình ta có: ${{n}_{Ca{{\left( OH \right)}_{2}}}}={{n}_{CaO}}=0,3\left( mol \right)$

${{C}_{{{M}_{Ca{{\left( OH \right)}_{2}}}}}}=\frac{{{n}_{Ca{{\left( OH \right)}_{2}}}}}{{{V}_{Ca{{\left( OH \right)}_{2}}}}}=\frac{0,3}{0,2}=1,5\left( M \right)$

$\Rightarrow $Chọn đáp án B