#### Answer

$3520\sqrt{2}y^{3}$

#### Work Step by Step

We find the term with $y^3$ in the expansion of $(\sqrt{2}+y)^{12}$ by using Binomial Theorem coefficients:
$\left(\begin{array}{l}
12\\9\end{array}\right)(\sqrt{2})^{9}y^{3}=220*(\sqrt{2})^8\sqrt{2}y^3=220*16\sqrt{2}y^3=3520\sqrt{2}y^{3}$