Answer
$^{21}_{9}F\rightarrow\,^{21}_{10}Ne+\,\,_{-1}^{0}e$
Work Step by Step
Let the missing nucleus be $^{A}_{Z}X$
Then, we have
$ ^{21}_{9}F\rightarrow\,^{A}_{Z}X+\,\,_{-1}^{0}e$
Since sum of the charges and masses should be equal on both the sides, we get
$21=A+0$ and $9= Z+(-1)$
$\implies A= 21$ and $Z= 10$.
The nucleus with $Z=10$ is that of $Ne$.
The missing nucleus is $^{21}_{10}Ne$.
The nuclear equation is
$^{21}_{9}F\rightarrow\,^{21}_{10}Ne+\,\,_{-1}^{0}e$
