Answer
$ k\approx 0.071$
Work Step by Step
Since the exponential decay is given by $$ P(t)=P_0 e^{-kt}, \quad k\gt 0$$ then we have
$$ P(t)=10 e^{-kt}.$$
At $ t=17$, $ P(17)=3$, so:
$$ P(17)=10 e^{-17k}=3\Longrightarrow k=-\frac{\ln(3/10)}{17}\approx 0.071.$$