Answer
$^{212}_{84}Po$
Work Step by Step
in $\beta^{-}$ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A}_{Z+1}D+^{0}_{-1}e$
Here $P$ is parent nucleus, $D$ is daughter nucleus and $^{0}_{-1}e$ is positron or $\beta^{-}$ particle
$^{208}_{81}Tl$ $\longrightarrow$ $^{208}_{82}Pb+^{0}_{-1}e$
so daughter of this decay is $^{208}_{82}Pb$
$^{208}_{82}Pb$ is also daughter of another $\alpha$ decay
in $\alpha $ deacy
$^{A}_{Z}P$ $\longrightarrow$ $^{A-4}_{Z-2}D+^{4}_{2}He$
Here $P$ is parent nucleus,$D$ is daughter nucleus and $^{0}_{+1}e$ is positron or $\beta^{+}$ particle
so if daughter is given then parent will have two more proton and in total four more nucleon.
so $^{212}_{84}Po$ $\longrightarrow$ $^{208}_{82}Pb+^{4}_{2}He$
