Answer
Please see the work below.
Work Step by Step
Lets say X be the product nucleus then the required equation is
$Sc^{41}_{21}\rightarrow X^{A}_{Z}+e ^{0}_{1}$
We obtain from the superscripts $41=A+0$ so $A=41$
From the subscripts we obtain $21=Z+1$ so $z=20$
Thus the product of the reaction is $X^{41}_{20}$ which shows its calcium with symbol Ca.
Hence we obtain the nuclear equation
$Sc^{41}_{21}\rightarrow O^{41}_{20}+e^{0}_{1}$
