Initially at rest, a uranium-238 nucleus emits an alpha particle...

By the law of conservation of linear momentum, the initial momentum of the system is zero.

Therefore, the magnitude of the momentum of the residual thorium nucleus is equal to the magnitude of the momentum of the alpha particle.

$m_{Th} v_{Th} = m_{\alpha} v_{\alpha}$

Given that the mass of a nucleus is proportional to its mass number, $m_{Th} \propto 234$ and $m_{\alpha} \propto 4$.

$234 \times v_{Th} = 4 \times 1.4 \times 10^7$

$v_{Th} = \dfrac{4 \times 1.4 \times 10^7}{234}$

$v_{Th} = \dfrac{5.6 \times 10^7}{234}$

$v_{Th} \approx 2.39 \times 10^5 \text{ m/s} \approx 2.4 \times 10^5 \text{ m/s}$