In a biprism experiment a steady interference pattern is observed on the screen using a light of wavelength $5000 Å$. Without disturbing the set up of the experiment, the source of light is replaced by a source of wavelength $6400 Å$. The fringe width will

Fringe width in a Fresnel biprism is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is wavelength, $D$ is source-screen distance, and $d$ is virtual source separation.
Since $D$ and $d$ remain unchanged, the fringe width is directly proportional to wavelength: $\beta \propto \lambda$.

With initial wavelength $\lambda_1 = 5000 \mathring{A}$ increasing to $\lambda_2 = 6400 \mathring{A}$, the fringe width ratio becomes $\frac{\beta_2}{\beta_1} = \frac{6400}{5000} = 1.28$.

The percentage change in fringe width is $(1.28 - 1) \times 100\% = 28\%$.
Since the ratio exceeds unity, the fringe width increases by $28\%$.

The correct answer is $\boxed{\text{D}}$.