Answer
After $0.585$ days, or over half a day, the radon would fall to $90\%$ of the original value.
Work Step by Step
The decay equation for radon-222 gas is $$y=y_0e^{-0.18t}$$
For the radon to fall to $90\%$ of the original value, it means $y=90\%y_0=0.9y_0$
To find $t$, substitute $y=0.9y_0$ back to the equation:
$$0.9y_0=y_0e^{-0.18t}$$ $$e^{-0.18t}=0.9$$
Take the natural logarithm of both sides:
$$-0.18t=\ln0.9$$ $$t=-\frac{\ln0.9}{0.18}\approx0.585(days)$$
So after $0.585$ days, or over half a day, the radon would fall to $90\%$ of the original value.
