Answer
$$14$$
Work Step by Step
$v(t)=\dfrac{dr}{dt}=-3 \sin t i +3 \cos t j +3\sqrt t k$
and $|v(t)|=\sqrt {(-3 \sin t)^2 +(3 \cos t)^2 +(3\sqrt t)^2}=3 \sqrt {1+t}$
$Arc \ Length ( L)=\int_a^b |v| \ dt \\=\int_0^3 3 \sqrt {1+t} \ dt \\=[2(1+t)^{3/2}]_0^{3} \\=2(1+3)^{3/2} -2 \\ =14$
