A parallel plate capacitor with area 200 cm$$^2$$ and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is $$25 \times 10^{-6}$$ N, the value of V is approximately: $$\left(\varepsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}\right)$$

First, let us convert every given quantity to SI units, because all standard electrostatic formulas are written in metres, newtons, volts, etc.

The plate area is given as 200 cm$$^2$$. We know that $$1\ \text{cm} = 10^{-2}\ \text{m}$$, so

$$200\ \text{cm}^2 = 200 \times (10^{-2}\ \text{m})^2 = 200 \times 10^{-4}\ \text{m}^2 = 2 \times 10^{-2}\ \text{m}^2.$$

The separation between the plates is 1.5 cm. Converting centimetres to metres gives

$$d = 1.5\ \text{cm} = 1.5 \times 10^{-2}\ \text{m} = 0.015\ \text{m}.$$

The attractive force between the plates is given as $$25 \times 10^{-6}\ \text{N}$$, which we can write in scientific notation as

$$F = 2.5 \times 10^{-5}\ \text{N}.$$

Now we need a relation connecting this force to the potential difference V. For a parallel-plate capacitor:

Capacitance $$C = \dfrac{\varepsilon_0 A}{d}.$$

Energy stored $$U = \dfrac{1}{2} C V^2.$$

The force of attraction is obtained by differentiating the energy with respect to distance while keeping V constant. The well-known result is

$$F = \dfrac{1}{2}\,\varepsilon_0\,A\,\left(\dfrac{V}{d}\right)^2.$$

We need V, so we solve the above expression for V step by step. Starting with

$$F = \dfrac{1}{2}\,\varepsilon_0\,A\,\dfrac{V^2}{d^2},$$

multiply both sides by 2 to clear the fraction:

$$2F = \varepsilon_0\,A\,\dfrac{V^2}{d^2}.$$

Now isolate $$V^2$$ by multiplying both sides by $$d^2$$ and then dividing by $$\varepsilon_0 A$$:

$$V^2 = \dfrac{2F\,d^2}{\varepsilon_0 A}.$$

We substitute every value we have already converted:

$$V^2 = \dfrac{2 \times (2.5 \times 10^{-5}\ \text{N}) \times (0.015\ \text{m})^2}{(8.85 \times 10^{-12}\ \text{C}^2/\text{N·m}^2)\times(2 \times 10^{-2}\ \text{m}^2)}.$$

First calculate $$d^2$$:

$$d^2 = (0.015\ \text{m})^2 = 0.000225\ \text{m}^2 = 2.25 \times 10^{-4}\ \text{m}^2.$$

Next, compute the numerator:

$$2F\,d^2 = 2 \times (2.5 \times 10^{-5}) \times (2.25 \times 10^{-4})$$

$$= 5.0 \times 10^{-5} \times 2.25 \times 10^{-4}$$

$$= 11.25 \times 10^{-9} = 1.125 \times 10^{-8}.$$

Now compute the denominator:

$$\varepsilon_0 A = (8.85 \times 10^{-12}) \times (2 \times 10^{-2})$$

$$= 17.7 \times 10^{-14} = 1.77 \times 10^{-13}.$$

Putting these results into the expression for $$V^2$$ gives

$$V^2 = \dfrac{1.125 \times 10^{-8}}{1.77 \times 10^{-13}}.$$

Divide the significant figures and handle the powers of ten separately:

Numerical part: $$\dfrac{1.125}{1.77} \approx 0.636.$$

Powers of ten: $$10^{-8} / 10^{-13} = 10^{5}.$$

So

$$V^2 \approx 0.636 \times 10^{5} = 6.36 \times 10^{4}.$$

To find V, take the square root:

$$V = \sqrt{6.36 \times 10^{4}} = \sqrt{6.36}\,\sqrt{10^{4}}.$$

We know $$\sqrt{10^{4}} = 10^{2} = 100,$$ and $$\sqrt{6.36} \approx 2.523.$$

Therefore

$$V \approx 2.523 \times 100\ \text{V} \approx 252\ \text{V}.$$

This value is closest to 250 V among the given options.

Hence, the correct answer is Option C.