Answer
$20$
Work Step by Step
Consider $I= \iint_{R} f(x,y) dA$
or, $= \int_{0}^{2} \int_{0}^{2-x} (12-3y^2) dy dx$
or, $= \int_{0}^{2}[12y-y^3]_0^{2-x} dx$
or, $=\int_{0}^{2} (24-12x-(2-x)^3) dx$
Thus, we have $I=[24 x-6x^2+ \dfrac{(2-x)^4}{4}]_0^2=20$