Answer
Please see the work below.
Work Step by Step
Lets say X be the product nucleus then the required equation is
$Ac^{227}_{84}\rightarrow X^{A}_{Z}+He ^{4}_{2}$
We obtain from the superscripts $227=A+4$ so $A=223$
From the subscripts we obtain $89=Z+2$ so $z=87$
Thus the product of the reaction is $X^{223}_{87}$ which shows its francium with symbol Fr.
Hence we obtain the nuclear equation
$Ac^{227}_{84}\rightarrow Fr^{223}_{87}+He^{4}_{2}$
