Answer
$^{137}_{55}Cs\rightarrow \,^{0}_{-1}e+\,^{137}_{56}Ba$
Work Step by Step
The beta particle is nothing but an electron.
In the case of beta decay, the new nuclide produced along with the emission of beta particle has the same mass number but the atomic number is increased by 1.
That is, $A=137$ and $Z=\text{(Z of Cs)}+1=55+1=56$
The nuclide with mass number 137 and atomic number 56 is $\,^{137}_{56}Ba$.
Therefore, the balanced nuclear equation is
$^{137}_{55}Cs\rightarrow \,^{0}_{-1}e+\,^{137}_{56}Ba$
