Explanation

(a) $\left| \frac{dN}{dt} \right|$= |Activity of radioactive substance|
$\ \ = \ \lambda N\ = \ \lambda N_{0}e^{- \lambda t\ \ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because\ N\ = \ N_{0}e^{- \lambda t})$
Taking log both sides
$\ln\left| \frac{dN}{dt} \right| = \ \ln(\ \lambda N_{0}) - \lambda t$
Hence, ln$\left| \frac{dN}{dt} \right|$ $versus$ t graph is a straight line with slope-$\lambda$.
From the graph we can see that,
$\lambda = \frac{1}{2} = 0.5\ {yr}^{- 1}$
Now applying the equation
$N\ = \ N_{0}\ e^{- \lambda t} = N_{0}\ e^{- 0.5\ \times \ 4.16}$
$\ \ \ \ \ \ = N_{0}e^{- 2.08}\ = \ 0.125\ N_{0}$
$\ \ \ \ \ \ \ \ \ \ = \ \frac{N_{0}}{8}$
$ie,\ $nuclei decreases by a factor of 8.
Hence the answer is 8.