1. Given data

• Wavelength: $\lambda = 6000\,\mathring{A} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$
• Slit width: $a = 0.01\,\text{mm} = 10^{-2} \times 10^{-3}\,\text{m} = 10^{-5}\,\text{m}$
• Focal length of lens: $f = 20\,\text{cm} = 0.2\,\text{m}$

2. Formula for central maximum width in Fraunhofer single-slit diffraction

At the focal plane of a convex lens, the linear width of the central maximum is

$W = 2\frac{f\lambda}{a}$

3. Substitute the values

$W = 2 \cdot \frac{(0.2)(6 \times 10^{-7})}{10^{-5}}$

First calculate inside the fraction:

$\frac{(0.2)(6 \times 10^{-7})}{10^{-5}} = 0.2 \times 6 \times 10^{-2} = 1.2 \times 10^{-2}\,\text{m}$

Now multiply by $2$:

$W = 2 \times 1.2 \times 10^{-2} = 2.4 \times 10^{-2}\,\text{m}$

4. Convert to mm

$2.4 \times 10^{-2}\,\text{m} = 24\,\text{mm}$

5. Match with options

$24\,\text{mm}$ corresponds to Option B.

Therefore, the linear width of the central maximum is $\boxed{24\,\text{mm}}$.