Answer
$E_1 = 4.25MeV$
Work Step by Step
When the decay releases $alpha$ particle, the nuclear reaction is written $^{238}U \longrightarrow ^{234}Th + ^{4}He $
The energy released is
$E_1 = (m_U - m_{He} - m_{Th}) c^2$
$E_1 = (238.05079u - 4.00260 u - 234.04363u) ( 931.5 MeV / u)$
$E_1 = 4.25MeV$
