$1) BaO + H_2O \to Ba(OH)_2$
$2) n_{Ba(OH)_2} = n_{BaO} = \dfrac{15,3}{153} = 0,1(mol)$
$\Rightarrow C_{M_{Ba(OH)_2}} = \dfrac{0,1}{0,5} = 0,2M$
$3) Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,2(mol)$
$V_{dd\ HCl} = \dfrac{0,2}{2} = 0,1(lít)$