Answer
$t = 59.5 d$
Work Step by Step
We use $R = R_0 e^{-\lambda t}$ and solve for t
$t = \frac{1}{\lambda} ln \frac{R_0}{R}$
Where
$\lambda = \frac{ln 2 }{t_{1/2}} $
$ \frac{1}{\lambda} = \frac{14.28 d}{ln 2}$
$ \frac{1}{\lambda} = \frac{14.28 y}{0.693}$
$ \frac{1}{\lambda} = 20.6 d $
So,
$t =(20.6 d) ln [\frac{3050}{170}]$
$t = 59.5 d$
