$\int_{1}^{2}\int_{0}^{x^{2}} x.dy.dx$ (this can be written as)
$\int_{1}^{2}\int_{0}^{x^{2}} 1 . x.dy.dx$
first integrating with respect to y
$\Rightarrow \int_{1}^{2}\left [ y \right ]_{0}^{x^2} x.dx$
$\Rightarrow \int_{1}^{2} x ^{2}.x.dx$
$\Rightarrow \int_{1}^{2} x^3.dx$
$\Rightarrow \left [ x^4/4 \right ]_{1}^{2}$
$\Rightarrow 16/4 -1/4$
$\Rightarrow 15/4$
hence answer is option A.