1. Given:

   • Frequency of incident light: $f = 5 \times 10^{14}\,\text{Hz}$
   • Refractive index of medium: $n = 2$
   • Speed of light in air (approximately vacuum): $c = 3 \times 10^8\,\text{m/s}$

2. Key concept: When light enters another medium, its frequency remains unchanged.

3. Speed of light in the medium: $v = \frac{c}{n} = \frac{3 \times 10^8}{2} = 1.5 \times 10^8\,\text{m/s}$

4. Wavelength in the medium: Using $\lambda = \frac{v}{f}$ we get $\lambda = \frac{1.5 \times 10^8}{5 \times 10^{14}}$

5. Simplify: $\lambda = 0.3 \times 10^{-6}\,\text{m} = 3 \times 10^{-7}\,\text{m}$

6. Convert to nm: $3 \times 10^{-7}\,\text{m} = 300\,\text{nm}$

7. Option check:

   • A: $400\,\text{nm}$ ❌
   • B: $300\,\text{nm}$ ✅
   • C: $600\,\text{nm}$ ❌
   • D: $500\,\text{nm}$ ❌

Therefore, the correct answer is: $\boxed{300\,\text{nm}}$