I had a go. If you want to apply Stokes' theorem you first need to decide whether it's easier to compute $$ \int \int F \cdot dS$$ in terms of $$ \oint F \cdot d\vec{r}$$ or the other way around.
Here $dS$ is the normal vector of your surface and $\vec{r}$ is the parametrisation of $C$, the boundary curve of your surface. In your homework, I decided that computing $ \oint F \cdot d\vec{r}$ was easier.
First you parameterise $C$ as follows: $$ \vec{r}(t) = \Big ( \begin{array}{c} 3 \cos t \\ 3 \sin t \\ 0 \end{array} \Big )$$
How did I get this parameterisation? The boundary circle is in the $xy$ plane, so $z = 0$. The other two are just the standard parameterisation of a circle of radius $3$.
Now we're good to go:
$$\begin{align} \int \int F \cdot dS = \oint_C F \cdot d \vec{r} = \int_0^{2\pi} F(r(t)) \cdot \vec{r} dt = \int_0^{2 \pi} \Big ( \begin{array}{c} 0 \\ 15 \cos t \\ 6 \sin t \end{array} \Big ) \cdot \Big ( \begin{array}{c} 3 \cos t \\ 3 \sin t \\ 0 \end{array} \Big ) dt = 45 \int_0^{2 \pi} \cos t \sin t dt = 0 \end{align}$$
Where I used integration by parts in the last step. Hope this helps.