When a uranium isotope ${}_{92}^{235}U$ is bombarded with a neutron, it generates ${}_{36}^{89}Kr$, three neutrons and :

${}_{92}{U}^{235}+{}_o{n}^1\to {}_Q{X}^P{+}_{36}K{r}^{89}+3{}_o{n}^1+Q(energy)$

$\sum$ Atomic number on $LHS=89+0=89+3$

$\sum$ Atomic number on $RHS=Q+36+3\times 0$

$\therefore LHS=RHS$

$Q+36=92$

$Q=56$

also, $\sum$ atomic mass number on 

$LHS=235+1=236$

$\sum$ Atomic mass number on 

$RHS=P+89+3\times 1$

$\therefore LHS=RHS$

$P+92=236$

$P=144$

So, the element is ${}_{56}B{a}^{144}$