= Solution
Use constant component <masses> and a circular <Keplerian orbit>. Then $J=\mu\sqrt{GMA}$ gives $\dot J/J=\dot A/(2A)$, while $P=2\pi A^{3/2}/\sqrt{GM}$ gives $\dot P/P=3\dot A/(2A)=3\dot J/J$. Dividing the supplied loss law by $J$ therefore yields
$$
\boxed{\frac{\dot P}{P}=-\frac{96}{5}\frac{G^3\mu M^2}{c^5A^4}},\qquad
\dot A=-\frac{64}{5}\frac{G^3\mu M^2}{c^5A^3}.
$$
The positive instantaneous <orbital decay timescale> is $\tau_P=P/|\dot P|$, since $\dot P<0$. For $t_G=10^{10}$ yr, the requested prescription gives
$$
A_{\max}=\left(\frac{96G^3\mu M^2t_G}{5c^5}\right)^{1/4},\qquad
P_{\max}=\frac{2\pi A_{\max}^{3/2}}{\sqrt{GM}}.
$$
It is important to distinguish this prescription from the integrated <gravitational-wave inspiral time>. Integrating $\dot A=-C/A^3$ gives
$$
\boxed{t_{\rm merge}=\frac{5c^5A^4}{256G^3\mu M^2}=\frac38\tau_P}.
$$
Finite contact radii supply a small lower-limit correction when the initial separation is much larger than the compact objects. Thus the periods below satisfy the printed instantaneous-timescale criterion and have a formal merger time $3.75$ Gyr. If instead the requirement is literally $t_{\rm merge}=10$ Gyr, every period limit is larger by $(8/3)^{3/8}=1.445$. Eccentric systems require a different decay calculation.
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