Use the mass defect formula for alpha decay

For the decay $^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He} + Q,$ the energy released is $Q = \Delta m\,c^2 = \left[m(^{238}\mathrm{U}) - m(^{234}\mathrm{Th}) - m(^{4}\mathrm{He})\right]c^2.$

Substitute the given masses

$m(^{238}\mathrm{U}) = 238.05060\,u$ $m(^{234}\mathrm{Th}) = 234.04360\,u$ $m(^{4}\mathrm{He}) = 4.00260\,u$

So, $\Delta m = 238.05060 - (234.04360 + 4.00260)$

Calculate the mass defect

First add the daughter nucleus and alpha particle masses: $234.04360 + 4.00260 = 238.04620\,u$

Therefore, $\Delta m = 238.05060 - 238.04620 = 0.00440\,u$

Convert mass defect into energy

Given $1u = 931.5\,\text{MeV}/c^2$

Hence, $Q = 0.00440 \times 931.5\,\text{MeV}$

$Q = 4.0986\,\text{MeV}$

Final integer answer

$Q \approx 4.1\,\text{MeV}$

Since the question asks for an integer, $\boxed{4}$