$BaO+{{H}_{2}}O\to Ba{{\left( OH \right)}_{2}}$

0,03           $\leftarrow $   0,03   (mol)

${{n}_{Ba{{\left( OH \right)}_{2}}}}={{V}_{Ba{{\left( OH \right)}_{2}}}}.{{C}_{{{M}_{Ba{{\left( OH \right)}_{2}}}}}}=0,3.0,1=0,03\left( mol \right)$

Theo phương trình ta có: ${{n}_{BaO}}={{n}_{Ba{{\left( OH \right)}_{2}}}}=0,03\left( mol \right)$

$\Rightarrow {{m}_{BaO}}={{n}_{BaO}}.{{M}_{BaO}}=0,03.153=4,59\left( g \right)$

$\Rightarrow $Chọn đáp án B