Friedmann equation

The equation becomes $\dot{a} \; = \; \sqrt{\frac{(1-\Omega)+\Omega a^3}{a}}$ and so $t-t_0 \; = \; \int_0^a\sqrt{\frac{u}{(1-\Omega)+\Omega u^3}}\,du \; = \; \tfrac23\int_0^{a\sqrt{a}} \frac{dv}{\sqrt{(1-\Omega) + \Omega v^2}} \; = \; \frac{2}{3\sqrt{\Omega}}\sinh^{-1}\sqrt{\frac{\Omega a^3}{1-\Omega}}$ so that $\tfrac32(t_1 - t_0)\sqrt{\Omega} \; = \; \sinh^{-1}\sqrt{\frac{\Omega}{1-\Omega}}$ If $A = a\big(\tfrac12(t_0+t_1)\big)$ , then $\frac{\Omega A^3}{1-\Omega} \; = \; \sinh^2\left(\tfrac34(t_1-t_0)\sqrt{\Omega}\right) \; = \; \sinh^2\left(\frac12\sinh^{-1}\sqrt{\frac{\Omega}{1-\Omega}}\right) \; = \; \frac12\left[\cosh\left(\sinh^{-1}\sqrt{\frac{\Omega}{1-\Omega}}\right) - 1 \right] \; = \; \frac{1 - \sqrt{1-\Omega}}{2\sqrt{1-\Omega}}$ and hence $A^3 \; = \; \frac{\sqrt{1-\Omega}(1 - \sqrt{1-\Omega})}{2\Omega}$ With $\Omega = 0.68$ we obtain $A^3 \; = \; \tfrac{1}{17}(5\sqrt{2}-4)$ and hence the answer is $\boxed{0.5653015309}$ . The fourth decimal place in the official answer is incorrect.

See my solution to this problem: https://brilliant.org/problems/after-the-big-bang/?ref_id=1555615 . As pointed out by Mark Hennings, the solution was off by 0.0001, because of a round-off error in an intermediary step.