Answer
$$\dfrac{17 \sqrt {17}-1}{4}$$
Work Step by Step
Since, $\vec{r} (x,z) =x i+x^2 j+zk$
Now, $\vec{r_x} \times \vec{r_z}=2x i \ -\ j +0 \ k$
$\implies |\vec{r_x} \times \vec{r_z}| =\sqrt {4x^2+1}$
$I=\int_{0}^2 \int_{0}^3 6 (x,y,z) \ d \theta =\int_{0}^2 \int_{0}^3 x \sqrt {4x^2+1} \ dz \ dx$
or, $=\int_0^2 (3x \sqrt {4x^2+1} ) \ dx$
Set $4x^2+1 u \implies du=8x dx$
Now,
$$I=\dfrac{3}{8} \int_{1}^{17} \sqrt u du \\=\dfrac{3}{8} \times [\dfrac{2}{3} \times (u)^{3/2}]_{1}^{17} \\=\dfrac{17 \sqrt {17}-1}{4}$$
