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

In the nuclear fission reaction:
$${}_{92}^{235}\text{U}{+}_0^1\text{n}{\to }_{36}^{89}\text{Kr}+3_0^1\text{n}{+}_Z^A\text{X}$$
Conservation of mass number: $235+1=89+3(1)+A\Rightarrow 236=92+A\Rightarrow A=144$.Conservation of atomic number: $92+0=36+3(0)+Z\Rightarrow 92=36+Z\Rightarrow Z=56$.The element with $Z=56$ is Barium. Thus, the product is ${}_{56}^{144}\text{Ba}$.