Using the remainder theorem, find the remainder,

$p(x)=6 x^{3}+13 x^{2}+3$

$g(x)=3 x+2=3\left(x+\frac{2}{3}\right)=3\left[x-\left(-\frac{2}{3}\right)\right]$

By remainder theorem, when $p(x)$ is divided by $(3 x+2)$, then the remainder $=p\left(-\frac{2}{3}\right)$.

Putting $x=-\frac{2}{3}$ in $p(x)$, we get

$p\left(-\frac{2}{3}\right)=6 \times\left(-\frac{2}{3}\right)^{3}+13 \times\left(-\frac{2}{3}\right)^{2}+3=-\frac{16}{9}+\frac{52}{9}+3=\frac{-16+52+27}{9}=\frac{63}{9}=7$

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.