Answer
$^{20}_{9}F\rightarrow \,\,^{0}_{-1}e+\,\,^{20}_{10}Ne$
Work Step by Step
There are excess neutrons in $\,^{20}_{9}F$, so beta emission is probable.
In the case of $\beta$ emission, mass number does not change, but the atomic number increases by one unit. Therefore, the resulting nuclide is one with the mass number 20 and atomic number=$9+1=10$. $Ne$(neon) is the element with atomic number 10.
Thus, we can write the nuclear equation for decay as
$^{20}_{9}F\rightarrow \,\,^{0}_{-1}e+\,\,^{20}_{10}Ne$
