Answer
173.3 MeV.
Work Step by Step
Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_{n}+m(^{235}_{92}U)-m(^{141}_{56}Ba)-m(^{92}_{36}Kr)-3m_{n} \right)c^2$$
$$ =\left(1.008665u+235.043930u-140.914411u-91.926156u-3(1.008665u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=173.3MeV$$
The Q-value is positive. The fission reaction is exothermic, and releases 173.3 MeV of energy.