When exponent is $n$ then total number of terms are $n+1$.
So, total number of terms in $(2+3 x) 4=5$
Middle term is $3 rd . \Rightarrow T_{3}={ }^{4} C_{2}(2)^{2} \cdot(3 x)^{2}$
$=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2} \times 4 \times 9 x^{2}$
$=216\, x^{2}$