Answer
The work done by the field along the given path is $\frac{9}{2}{\pi ^2}$.
Work Step by Step
We obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\cos t,\sin t,t} \right)$
${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,1} \right)$
The work done by the field:
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{3\pi } {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^{3\pi } \left( {\cos t,\sin t,t} \right)\cdot\left( { - \sin t,\cos t,1} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^{3\pi } t{\rm{d}}t = \frac{1}{2}{t^2}|_0^{3\pi } = \frac{9}{2}{\pi ^2}$
So, the work done by the field along the given path is $\frac{9}{2}{\pi ^2}$.