Answer
$^{47}_{20}Ca\rightarrow \,^{0}_{-1}e+\,^{47}_{21}Sc$
Work Step by Step
In the case of beta decay, the new nuclide produced along with the emission of beta particle has same mass number but atomic number is increased by 1.
That is, $A=47$ and Z= Z of Ca+1=20+1=21.
The nuclide with mass number 47 and atomic number 21 is $\,^{47}_{21}Sc$.
The beta particle is nothing but an electron.
Therefore, the balanced nuclear equation is
$^{47}_{20}Ca\rightarrow \,^{0}_{-1}e+\,^{47}_{21}Sc$
