A graph is plotted between the natural logarithm of the nuclear...

The relation between the nuclear radius $R$ and the mass number $A$ is given by:
$R = R_0 A^{1/3}$

Taking the natural logarithm on both sides, we get:
$\ln R = \ln R_0 + \dfrac{1}{3} \ln A$

Comparing this equation with the standard equation of a straight line, $y = mx + c$, where $y = \ln R$ and $x = \ln A$, we find:
Slope ($m$) = $\dfrac{1}{3}$
y-intercept ($c$) = $\ln R_0$

Therefore, the slope of the graph is $\dfrac{1}{3}$.

Answer: $\dfrac{1}{3}$