${}^{235}_{92}U$ nucleus absorbs a neutron and disintegrates into ${}^{139}_{54}Xe$, ${}^{94}_{38}Sr$ and x. So, what will be the product x : -

Fission reaction: ${}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{139}_{54}Xe + {}^{94}_{38}Sr + x$
Atomic numbers: left side $92 + 0 = 92$; right side $54 + 38 = 92$. So x carries no charge (Z = 0).
Mass numbers: left side $235 + 1 = 236$; right side $139 + 94 = 233$. So x has a total mass number of $236 - 233 = 3$.
Particles with Z = 0 and A = 1 are neutrons, so x = 3 neutrons: ${}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{139}_{54}Xe + {}^{94}_{38}Sr + 3\,{}^{1}_{0}n$
So the product x is 3 neutrons.
Note: the answer key of the source gives 2 neutrons; it is changed here because the mass numbers balance only with 3 neutrons.