Right option is (c) Width of central maximum decreases
For explanation I would say: As the whole apparatus is now immersed in water, the wavelength of the light will change.
λ’=\(\frac{\lambda}{\mu}\)
Therefore, as the refractive index of water is greater than the air, the wavelength of light will decrease.
Width of central maxima = \(\frac{2\lambda}{a}\)
Therefore, as the wavelength decreases, the width of the central maxima decreases.