Answer
$\beta = 0.99999915$
Work Step by Step
We can find $\gamma$:
$\Delta t = \gamma \Delta t_0$
$\gamma = \frac{\Delta t}{\Delta t_0}$
$\gamma = \frac{23,000~years}{30~years}$
$\gamma = \frac{2300}{3}$
We can find the speed parameter $\beta$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\sqrt{1-\beta^2} = \frac{1}{\gamma}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{(2300/3)^2}}$
$\beta = 0.99999915$
