The quarter-cylinder is divided into five surfaces.

We label $S_1$ and $S_2$ for the 'caps' $y = 0$ and $y = 8$; the side and bottom, $x = 0$ and $z = 0$ is called $S_2$ and $S_4$ and $S_5$ denotes the curved cylinder. The normals are as follows:
\begin{align*}
S_1: \quad \mathbf{n} &= [0, -1, 0] \\
S_2: \quad \mathbf{n} &= [0, 1, 0] \\
S_3: \quad \mathbf{n} &= [-1, 0, 0] \\
S_4: \quad \mathbf{n} &= [0, 0, -1] \\
S_5: \quad \mathbf{n} &= [\cos\theta, 0, \sin\theta].
\end{align*}

All the normals are obvious with the exception of the last one. The normal for the curved portion of the cylinder is simply a normal of a circle in the $xz$ plane (and with a zero $y$ component). The computation of all four integrals can be done as follows. We let $\mathbf{A} \cdot \mathbf{n} = \mathbf{F}(x,y,z)$. Then
\begin{align*}
\int_{S_1} \mathbf{A} \cdot \mathbf{n} \, \de{S_1} &= \int_{r=0}^{r=3}\int_{\theta=0}^{\theta=\pi/2} F(r\cos\theta,0, r\sin\theta) \, r \, \de{\theta}\de{r} = -18 \\
\int_{S_2} \mathbf{A} \cdot \mathbf{n} \, \de{S_2} &= \int_{r=0}^{r=3}\int_{\theta=0}^{\theta=\pi/2} F(r\cos\theta,8,r\sin\theta) \, r \, \de{\theta}\de{r} = 18 + 18\pi \\
\int_{S_3} \mathbf{A} \cdot \mathbf{n} \, \de{S_3} &= \int_{z=0}^{z=3}\int_{y=0}^{y=8} F(0,y,z) \, \de{y}\de{z} = -216 \\
\int_{S_4} \mathbf{A} \cdot \mathbf{n} \, \de{S_4} &= \int_{x=0}^{x=3}\int_{y=0}^{y=8} F(x,y,0) \, \de{y}\de{x} = 36 \\
\int_{S_5} \mathbf{A} \cdot \mathbf{n} \, \de{S_5} &= \int_{y=0}^{y=8}\int_{\theta=0}^{\theta=\pi/2} F(3\cos\theta, y, 3\sin\theta) \, 3 \, \de{\theta}\de{y} = 180.
\end{align*}

The total surface integral is then $-18 + [18 + 18\pi] - 216 + 36 + 180 = 18\pi$. Remember that when you do the computations, the first four integrals do not need a scaling factor (because the surfaces are planar), whereas the last one requires multiplication by the cylinder radius.