Write the fission reaction

Let the number of neutrons emitted be $x$:

${}_{92}^{235}U \rightarrow {}_{54}^{142}Xe + {}_{38}^{90}Sr + x\,{}_{0}^{1}n$

Check conservation of atomic number

Atomic number on LHS: $92$

Atomic number on RHS: $54 + 38 = 92$

So, atomic number is balanced.

Apply conservation of mass number

Mass number on LHS: $235$

Mass number on RHS: $142 + 90 + x = 232 + x$

Equating both sides: $235 = 232 + x$

$x = 3$

Conclusion

The number of neutrons emitted is:

$\boxed{3}$