A

Step 1: Identify the reaction: \( {}^{235}_{92}U + {}^{1}_{0}n \rightarrow {}^{95}_{42}Mo + {}^{139}_{57}La + 2 \times {}^{1}_{0}n \)
Step 2: Calculate the mass of reactants: \( m_{reactants} = m(^{235}U) + m(n) = 235.0439\,u + 1.0087\,u = 236.0526\,u \)
Step 3: Calculate the mass of products: \( m_{products} = m(^{95}Mo) + m(^{139}La) + 2 \times m(n) = 94.9056\,u + 138.9061\,u + 2 \times 1.0087\,u = 235.8295\,u \)
Step 4: Find the mass defect: \( \Delta m = m_{reactants} - m_{products} = 236.0526\,u - 235.8295\,u = 0.2231\,u \)
Step 5: Convert mass defect to energy: \( E = \Delta m \cdot c^2 = 0.2231\,u \times 931.5\,MeV/c^2/u = 208.37\,MeV \)
Therefore, the energy released by one \( {}^{235}_{92}U \) nucleus is approximately \( 208.37\,MeV \). Thus, the correct answer is A.