Answer
$k = 0.079~N/m$
Work Step by Step
We can assume that the period is 1.0 s. We can find the spring constant of the silk thread. Therefore;
$T = 2\pi~\sqrt{\frac{m}{k}}$
$k = \frac{(2\pi)^2~m}{T^2}$
$k = \frac{(2\pi)^2~(0.0020~kg)}{(1.0~s)^2}$
$k = 0.079~N/m$
