Answer
See the plot of the clothoid (figure attached) ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$, where
$x\left( t \right) = \mathop \smallint \limits_0^t \sin \frac{{{u^3}}}{3}{\rm{d}}u$, ${\ \ \ }$ $y\left( t \right) = \mathop \smallint \limits_0^t \cos \frac{{{u^3}}}{3}{\rm{d}}u$
Its curvature is $\kappa \left( t \right) = {t^2}$.
Work Step by Step
Using a computer algebra system, we plotted the clothoid ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$, where
$x\left( t \right) = \mathop \smallint \limits_0^t \sin \frac{{{u^3}}}{3}{\rm{d}}u$, ${\ \ \ }$ $y\left( t \right) = \mathop \smallint \limits_0^t \cos \frac{{{u^3}}}{3}{\rm{d}}u$
By the Fundamental Theorem of Calculus:
$x'\left( t \right) = \sin \frac{{{t^3}}}{3}$, ${\ \ \ }$ $y'\left( t \right) = \cos \frac{{{t^3}}}{3}$
The second derivatives are
$x{\rm{''}}\left( t \right) = {t^2}\cos \frac{{{t^3}}}{3}$, ${\ \ \ }$ $y{\rm{''}}\left( t \right) = - {t^2}\sin \frac{{{t^3}}}{3}$
Substituting these in Eq. (11) of Exercise 28:
$\kappa \left( t \right) = \frac{{\left| {x'\left( t \right)y{\rm{''}}\left( t \right) - x{\rm{''}}\left( t \right)y'\left( t \right)} \right|}}{{{{\left( {x'{{\left( t \right)}^2} + y'{{\left( t \right)}^2}} \right)}^{3/2}}}}$
gives
$\kappa \left( t \right) = \frac{{\left| { - {t^2}{{\sin }^2}\frac{{{t^3}}}{3} - {t^2}{{\cos }^2}\frac{{{t^3}}}{3}} \right|}}{{{{\left( {{{\sin }^2}\frac{{{t^3}}}{3} + {{\cos }^2}\frac{{{t^3}}}{3}} \right)}^{3/2}}}} = \frac{{\left| { - {t^2}\left( {{{\sin }^2}\frac{{{t^3}}}{3} + {{\cos }^2}\frac{{{t^3}}}{3}} \right)} \right|}}{{{{\left( {{{\sin }^2}\frac{{{t^3}}}{3} + {{\cos }^2}\frac{{{t^3}}}{3}} \right)}^{3/2}}}}$
Since ${\sin ^2}\frac{{{t^3}}}{3} + {\cos ^2}\frac{{{t^3}}}{3} = 1$, so $\kappa \left( t \right) = \left| { - {t^2}} \right| = {t^2}$.
