The angular width of fringes in Young's Double Slit Experiment is given by $\theta = \frac{\lambda}{d}$. When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light in the liquid becomes $\lambda' = \frac{\lambda}{\mu}$. Therefore, the new angular width $\theta'$ will be: $\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}$ Given, initial angular width $\theta = 0.2^{\circ}$ and refractive index of water $\mu = \frac{4}{3}$. Substituting the values: $\theta' = \frac{0.2^{\circ}}{4/3} = 0.2^{\circ} \times \frac{3}{4} = 0.15^{\circ}$. Thus, the new angular width of the first minima will be $0.15^{\circ}$.