Answer
$$
1.2 \mathrm{\ MeV}
$$
Work Step by Step
$\text { We solve } Q_{\text {eff }} \text { from } P=R Q_{\text {eff: }} $
$$
\begin{aligned} Q_{\text {eff }} &=\frac{P}{R}=\frac{m P T_{1 / 2}}{M \ln 2} \\ &=\frac{\left(90.0 \text { u) }\left(1.66 \times 10^{-27} \mathrm{kg} / \mathrm{u}\right)(0.93 \mathrm{W})(29 \mathrm{y})\left(3.15 \times 10^{7} \mathrm{s} / \mathrm{y}\right)\right.}{\left(1.00 \times 10^{-3} \mathrm{kg}\right)(\ln 2)\left(1.60 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)} \\ &=1.2 \mathrm{\ MeV} \end{aligned}
$$
