The correct answer is $\boxed{\text{B) }0.67}$.
The formula for the arc length of a curve $y=f(x)$ over the interval $[a,b]$ is:
$$L = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$
In this case, we have $y=\frac{2}{3}x^{3/2}$ and $a=0$ and $b=1$. Substituting these values into the formula, we get:
$$L = \int_0^1 \sqrt{1+\left(\frac{d}{dx}\left(\frac{2}{3}x^{3/2}\right)\right)^2} dx = \int_0^1 \sqrt{1+\left(\frac{2}{3}\cdot\frac{3}{2}x^{^{-1/2}}\right)^2} dx = \int_0^1 \sqrt{1+\frac{4}{9}x^{-1}} dx$$
We can now evaluate this integral using the substitution $u=\frac{1}{x}$. This gives us:
$$L = \int_0^1 \sqrt{1+\frac{4}{9}x^{-1}} dx = \int_1^\infty \sqrt{1+\frac{4}{9}u^2} du = \left.\frac{2}{3}u^{3/2}\sqrt{1+\frac{4}{9}u^2}\right|_1^\infty = \frac{2}{3}\sqrt{1+\frac{4}{9}} – \frac{2}{3} = 0.67$$
Therefore, the length of the curve $y=\frac{2}{3}x^{3/2}$ between $x=0$ and $x=1$ is $\boxed{\text{B) }0.67}$.