You only made one computational error and a notational error.
Here is your solution (at least the first part that follows) re-written:
You are minimizing the cost of the fence, so you want to minimize the perimeter of the enclosure. The perimeter $P$
is
$$\tag{1}
P=2l+3w.
$$
where $l$ is the length and $w$ is the width of the enclosure. It is assumed the division of the original fence is made across a width (hence the $3w$).
We want to write $P$ in terms of one variable. For this we use the given information that the area is $1.5\cdot10^6\,\rm ft^2$. So $$lw= 1.5\cdot10^6\,\rm ft^2,$$ or,
$$\tag{2}
l={1.5\cdot10^6\over w}
$$
Substituting $(2)$ into $(1)$ gives
$$
P=2\bigl({1.5\cdot10^6\over w}\bigr) +3w
$$
or
$$
P(w)= {3\cdot10^6\over w} +3w
$$
(this is where you made your mistake).
We also need to find the appropriate range of values for $w$. Here, $w$ could be possibly any positive, finite length.
So, you want to find the minimum value of $P$ over the interval $(0,\infty)$.
Towards this end, evaluate
$$
P'(w)= - {3\cdot10^6\over w^2} +3.
$$
Set $P'(w)=0$ and find the solutions in the interval $(0,\infty)$:
$$
- {3\cdot10^6\over w^2} +3 = 0 \quad\iff\quad {3\cdot10^6 } =3w^2\quad\iff\quad w=10^3.
$$
So $w=1000\,\rm ft$ gives the only critical point of $P$ in the interval $(0,\infty)$.
We need to examine what goes on near the endpoints of our interval before declaring that $1000\,\rm ft$ gives the answer.
As you can easily convince yourself $\lim\limits_{w\rightarrow 0^+} P(w)$ and $\lim\limits_{w\rightarrow \infty} P(w)$ are both infinite. So the cost of the enclosure is minimized when $w=1000\,\rm ft$.
And to explicitly answer the question (don't forget to do this): "he can do this by taking the width to be $1000\,\rm ft$, the length to be ${1.5\cdot 10^6\over1000}=1500\,\rm ft$ and spliting the enclosure by constructing another width".
Let the two inside parallel walls each have length $x$. Let the sides of the rectangle perpendicular to these each have length $y$.
Then the total area enclosed is $xy$. The amount of fencing used is $4x+2y$. This is to be $900$, since it is clear that it is best to use up all the fencing.
So we want to maximize $xy$, under the constraint $4x+2y=900$.
Thus $y=450-2x$, and we want to maximize $x(450-2x)$.
Because of the physical situation, we need $x\ge 0$ and $y\ge 0$. This means $x\le 225$.
So mathematically, we want to minimize $f(x)=450x-2x^2$, where $0\le x\le 225$.
This can be done by standard tools, such as calculus or completing the square.
Best Answer
Hint: Draw a picture. Let the length of the "middle" fence be $x$. Write $x$ beside this length of fence. Write $x$ beside the two sides of the field that are parallel to the middle fence. Write $y$ beside the remaining two sides of the fenced region.
The total amount of fencing we use is $3x+2y$. This is what we want to minimize. Note that $xy=10^6$. It follows that the amount $F(x)$ of fencing that we use is given by $$F(x)=3x+\frac{2\times 10^6}{x}.$$ Minimize $F(x)$ using the usual techniques.