Answer
126.5 MeV.
Work Step by Step
Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_{n}+m(^{235}_{92}U)-m(^{88}_{38}Sr)-m(^{136}_{54}Xe)-12m_{n} \right)c^2$$
$$ =\left(1.008665u+235.043930u-87.905612u-135.907214u-12(1.008665u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=126.5MeV$$
The Q-value is positive. The fission reaction is exothermic, and releases 126.5 MeV of energy.