To find the new level of production when 25 more workers are employed, we need to integrate the given rate of change of production with respect to the number of workers over the interval from the current number of workers to the current number plus 25.

The rate of change of production is given by the differential equation:

$$\frac{\mathrm{d} P}{\mathrm{~d} x}=100-12 \sqrt{x}$$

To find the production function $$P(x)$$, we integrate the differential equation with respect to $$x$$:

$$P(x) = \int(100-12\sqrt{x})\,\mathrm{d}x$$

Integrating term by term, we get:

$$\begin{aligned} P(x) &= \int 100 \, \mathrm{d}x - \int 12\sqrt{x} \, \mathrm{d}x \\\\ &= 100x - 12 \int x^{\frac{1}{2}} \, \mathrm{d}x \\\\ &= 100x - 12 \cdot \frac{2}{3}x^{\frac{3}{2}} + C \\\\ &= 100x - 8x^{\frac{3}{2}} + C \end{aligned}$$

Where $$C$$ is the constant of integration.

We know that when the number of workers was 0, the firm was manufacturing 2000 items. We can use this information to find the constant $$C$$:

$$2000 = P(0) = 100 \cdot 0 - 8 \cdot 0^{\frac{3}{2}} + C$$ $$C = 2000$$

Now, the production function with the constant $$C$$ included is:

$$P(x) = 100x - 8x^{\frac{3}{2}} + 2000$$

The new level of production after employing 25 more workers is $$P(25)$$, provided that the initial number of workers was 0. If there was an initial number of workers $$x_0$$ where $$x_0 \ge 0$$, we need to evaluate the increase in production from $$x_0$$ to $$x_0 + 25$$.

However, without loss of generality and because no initial number of workers $$x_0$$ is provided, let's proceed by assuming the firm is starting from 0 workers. We add 25 workers to find:

$$\begin{aligned} P(25) &= 100 \cdot 25 - 8 \cdot 25^{\frac{3}{2}} + 2000 \\\\ &= 2500 - 8 \cdot (5^2)^{\frac{3}{2}} + 2000 \\\\ &= 2500 - 8 \cdot 5^3 + 2000 \\\\ &= 2500 - 8 \cdot 125 + 2000 \\\\ &= 2500 - 1000 + 2000 \\\\ &= 2500 + 1000 \\\\ &= 3500 \end{aligned}$$

Thus, the correct answer is 3500, which corresponds to Option D.

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