${}_{\mathrm{Z}}{}^{37}\mathrm{A}^{-}$ has 11.1% ore neutron than electrons. A stands for

Let ${\mathrm{e}}^{-}=\mathrm{x}\Rightarrow \mathrm{z}=\mathrm{x}-1$

and,  n = x + $\frac{11.1}{100}x=1.111x$

p + n = 56          

x -1 + 1.111 x = 37

x = $\frac{38}{2.111}=18$

z = x - 1 = 18 = 17

Hence, ${}_{17}{}^{37}\mathrm{A}$ is Cl.