Three times the first of three consecutive odd integers is $3$ more than twice the third. The third integer is:

Let the three consecutive odd integers be represented as:
* First integer $=x$
* Second integer $=x+2$
* Third integer $=x+4$

According to the given condition, three times the first integer is $3$ more than twice the third integer:

$$3x=2(x+4)+3$$

Now, expand and solve the linear equation:

$$3x=2x+8+3$$

$$3x=2x+11$$

$$3x-2x=11$$

$$x=11$$

The value of the first odd integer is $11$. We need to find the value of the third integer:

$$\text{Third integer}=x+4=11+4=15$$