The volume of the solid surrounded by the surface \[{\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right)^{\frac{2}{3}}} + {\left( {\frac{{\rm{y}}}{{\rm{b}}}} \right)^{\frac{2}{3}}} + {\left( {\frac{{\rm{z}}}{{\rm{c}}}} \right)^{\frac{2}{3}}} = 1\] is A. \[\frac{{4\pi {\rm{abc}}}}{{35}}\] B. \[\frac{{{\rm{abc}}}}{{35}}\] C. \[\frac{{2\pi {\rm{abc}}}}{{35}}\] D. \[\frac{{\pi {\rm{abc}}}}{{35}}\]

” option2=”\[\frac{{{\rm{abc}}}}{{35}}\]” option3=”\[\frac{{2\pi {\rm{abc}}}}{{35}}\]” option4=”\[\frac{{\pi {\rm{abc}}}}{{35}}\]” correct=”option1″]

The correct answer is $\boxed{\frac{{\pi {\rm{abc}}}}{{35}}}$.

The surface $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$ is an ellipsoid. The volume of an ellipsoid is given by the formula

$$V = \frac{4}{3} \pi abc \int_0^1 \int_0^1 \int_0^1 \sqrt{x^2/a^2 + y^2/b^2 + z^2/c^2} dx dy dz.$$

In this case, we have $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$, so we can simplify the integral to

$$V = \frac{4}{3} \pi abc \int_0^1 \int_0^1 \int_0^1 \sqrt{1} dx dy dz = \frac{4}{3} \pi abc.$$

The other options are incorrect because they do not take into account the fact that the surface is an ellipsoid.