Hurray 2016!

Let $T_{r}$ denote the co-efficient of $r^{th}$ term in the expansion of $\left(3x+2\right)^{2016}$
$T_{r+1} = \dbinom{2016}{r}3^{2016-r}2^{r}$

Replacing r by r-1,
$T_{r} = \dbinom{2017-r}{r-1}3^{2017-r}2^{r-1}$
$\dfrac{T_{r+1}}{T_{r}} = \dfrac{2(2017-r)}{3r} ...(1)$

Solving for $\dfrac{T_{r+1}}{T_{r}} \ge 1$ yields $r \le \dfrac{4034}{5}$
$\therefore r \le 806.8$

Replacing r by r+1 in $(1)$ ,

$\dfrac{T_{r+2}}{T_{r+1}} = \dfrac{2(2016-r)}{3(r+1)}$

Solving for $\dfrac{T_{r+2}}{T_{r+1}} \le 1$ yields $r \ge \dfrac{4029}{5}$
$\therefore r \ge 805.8$
The only integer satisfying both conditions is 806.
$\therefore r = 806$

Hence the greatest co-efficient term is $(r+1) = 806+1 = 807$ term.
Co-efficient of x in $807^{th}$ term $= 2016-806 = 1210$

General problem : For what power $x^k$ does the expansion of $(ax + b)^n$ have the largest coefficient?

Solution : The coefficient of $x^k$ is $C_k = \left(\begin{array}{c} n \\ k\end{array}\right) a^k b^{n-k} = \frac{n\cdots(n-k+1)}{k\cdots 1} a^k b^{n-k}.$ This coefficient is greater than the previous one iff $C_k > C_{k-1} \\ \frac{n\cdots(n-k+1)}{k\cdots 1} a^k b^{n-k} > \frac{n\cdots(n-k+2)}{(k-1)\cdots 1} a^{k-1} b^{n-k+1} \\ \frac{n-k+1}{k} > \frac b a \\ \frac{n+1}{k} > \frac{b+a}a \\ k < (n+1) \frac a{b+a}.$ The greatest coefficient corresponds to the greatest integer $k$ for which this is the case, so $k = \left\lfloor (n+1) \frac a{b+a}\right\rfloor.$

In this case: $k = \left\lfloor 2017\cdot \frac 3 5 \right\rfloor = \boxed{1210}.$

Interesting fact: If we divide the entire expression by $5^{2016}$ we get $(\tfrac35x + \tfrac25)^{2016}$ , in which the coefficient of $x_k$ corresponds to the probability of having $k$ successes in an $2016$ -times repeated experiment with success probability $p = \tfrac35$ . The "peak" of this distribution, which corresponds to the greatest exponent, is the expectation value of this distribution, $\mathbb E = np = 2016\cdot \tfrac35 \approx 1210.$ This is how I originally guessed the answer to this question. Of course, the derivation above is more "solid" than this intuitive approach.

There are nice solutions provided by both @Vighnesh Shenoy and @Manuel Kahayon. I only wanted to show the generalization to this problem by treating everything symbolically :

What is the exponent of $x$ that have the greatest coefficient in the expansion of $(ax+b)^{N}$ ?

The approach is not far from how you would solve the previous problem, you will arrive at

$\displaystyle \bigg \lfloor \frac{a(N+1)}{a+b}\bigg \rfloor$

What we need to be aware of is the sign of $a$ or $b$ . If exactly one of them is negative, we have to check if their corresponding index is even. If not, we have to check which neighboring coefficient is larger. In the case where we introduced negatives, it is necessary to change the expression to :

$\displaystyle \bigg \lfloor \frac{|a|(N+1)}{|a|+|b|}\bigg \rfloor$

Since the first term is $(3x)^{2016}$ , and the nth term is equal to $\frac{(n-1th term)(2018-n)(2)}{(3)(n-1)}$ , then as long as $\frac{(2018-n)(2)}{(3)(n-1)} \geq 1$ , then the nth term will be greater than the n-1th term. Equating this, we get

$\frac{(2018-n)(2)}{(3)(n-1)} \geq 1$

$807.8 \geq n$

Rounding off, the 807th term has the largest coefficient, meaning, the 807th term has $1210$ is the exponent of x in the 807th term.