A 2 meter long scale with least count of 0.2 cm is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at 80 cm mark and 1 m mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at 180 cm mark. The % error in the estimation of focal length is:

Object at 80 cm mark, lens at 100 cm mark, image at 180 cm mark. Least count = 0.2 cm.

Object distance: $$u = 80 - 100 = -20$$ cm (taking sign convention).

Image distance: $$v = 180 - 100 = 80$$ cm.

Using the lens formula: $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{80} - \dfrac{1}{-20} = \dfrac{1}{80} + \dfrac{1}{20} = \dfrac{1 + 4}{80} = \dfrac{5}{80} = \dfrac{1}{16}$$

So $$f = 16$$ cm.

Differentiating $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$:

$$\dfrac{\Delta f}{f^2} = \dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$$

The error in each measurement: $$\Delta u = \Delta v = 0.2$$ cm (least count).

$$\dfrac{\Delta f}{f^2} = \dfrac{0.2}{80^2} + \dfrac{0.2}{20^2} = \dfrac{0.2}{6400} + \dfrac{0.2}{400} = 0.00003125 + 0.0005 = 0.00053125$$

$$\Delta f = f^2 \times 0.00053125 = 256 \times 0.00053125 = 0.136$$ cm.

$$\% \text{ error} = \dfrac{\Delta f}{f} \times 100 = \dfrac{0.136}{16} \times 100 = 0.85\%$$

The answer is Option 1: $$\boxed{0.85}\%$$.