Answer
Please see the work below.
Work Step by Step
Lets say X be the product nucleus then the required equation is
$PO^{210}_{84}\rightarrow X^{A}_{Z}+He ^{4}_{2}$
We obtain from the superscripts $210=A+4$ so $A=206$
From the subscripts we obtain $84=Z+2$ so $z=82$
Thus the product of the reaction is $X^{206}_{82}$ which shows its Lead with symbol Pb.
Hence we obtain the nuclear equation
$Po^{210}_{84}\rightarrow Pb^{206}_{82}+He^{4}_{2}$
