Answer
$_{87}^{221}Fr$
Work Step by Step
$\alpha$-decay, in general, can be represented as
$_{Z}^{A}X\rightarrow_{Z-2}^{A-4}Y+ _{2}^{4}He$
Now, the nuclear equation can be written as
$ _{85+2}^{217+4}X\rightarrow_{85}^{217}At+ _{2}^{4}He$
The element X with atomic number 87 is Francium(Fr).
Therefore, the nuclear equation is
$ _{87}^{221}Fr\rightarrow_{85}^{217}At+ _{2}^{4}He$
