As a general rule in any decay sum of mass number $A$ and atomic number $Z$ must be same on both sides.
Let the daughter nucleus be ${ }_{ Z }^{ A } X$.
So, reaction can be shown as
${ }_{92}^{238} U \longrightarrow { }_{ Z }^{ A } X +{ }_{2}^{4} He$
From conservation of atomic mass
$ 238 = A +4$
$\Rightarrow A =234$
From conservation of atomic number
$ 92 = Z +2 $
$\Rightarrow Z =904$
So, the resultant nucleus is ${ }_{90}^{234} X$,
i.e., ${ }_{90}^{234} Th$.
Note: A nuclide below the stability line in uranium decay series disintegrates in such a way that its proton number decreases and its neutron to proton ratio increases. In heavy nuclides this can occur by alpha emission.