a,
$n_{BaO}=\frac{15,3}{153}=0,1 mol$
$BaO+H_2O\to Ba(OH)_2$
$\Rightarrow n_{Ba(OH)_2}=0,1 mol$
$C_{M_{Ba(OH)_2}}=\frac{0,1}{0,5}=0,2M$
b,
$Ba(OH)_2+2HCl\to BaCl_2+2H_2O$
$\Rightarrow n_{HCl}=0,1.2=0,2 mol$, $n_{BaCl_2}=0,1 mol$
$V_{HCl}=\frac{0,2}{2}=0,1l=100ml$
c,
$n_{BaCl_2}= n_{BaCl_2.2H_2O}=0,1 mol$
$\Rightarrow m_B=0,1.(208+18.2)=24,4g$