Let x equal the cost per unit for the housing project. Then the question stem gives:

$$f\cdot x=1,750,000$$

Statement 1

This tells us that we can build f/2 units for 1,750,000, where x = 250,000. In other words:

$$\frac{f}{2}\cdot250,000\ =\ 1,750,000$$

Solving for f gives 14 units. We can then plug into our initial equation:

$$14\cdot x=1,750,000$$

Solving for x gives $125,000 per unit. Sufficient.

Statement 2

This tells us that we can build f + 6 units for 1,750,000 at x - 3,750 per unit. In other words:

$$\left(f+6\right)\cdot\left(x-3,750\right)\ =\ 1,750,000$$

We can rearrange our original equation to get

$$f\ =\ \frac{1,750,000}{x}$$

Then we substitute that in for f
$$\left(\frac{1,750,000}{x}+\ 6\right)\left(x-3750\right)=1,750,000$$

At this point, we could expand, giving a quadratic equation that we can then factor, but we should recognize at this point that we CAN do that, hence we CAN solve for x, and statement 2 is sufficient. These numbers aren't pretty, so continuing to solve the quadratic is a waste of time.

Another trick we can use on statement 2: if we have n variables, we need n distinct equations to solve for them.

In statement 2, we have 2 variables (f and x) and 2 distinct equations (the one from the question stem and the one from statement 2). So we should know even without substituting in for f that we can solve for x using these two equations together.