Angle of deviation $ {{\delta }_{air}} $ is given by when the prism is in air $ {{\delta }_{air}}=(n-1)A=\left( \frac{3}{2}-1 \right)A=\frac{A}{2} $ Again when the prism is immered in water then angle of deviation $ {{\delta }_{water}} $ $ {{\delta }_{water}}=({}_{g}{{n}_{w}}-1)A $ Here $ {}_{g}{{n}_{w}} $ is the refractive index of glass with respect to water so, $ {}_{g}{{n}_{w}}=\frac{{{n}_{g}}}{{{n}_{w}}}=\frac{3\text{/}2}{4\text{/}3}=\frac{9}{8} $ So, $ {{\delta }_{water}}=\left( \frac{9}{8}-1 \right)A=\frac{1}{8}A $ Hence, required ratio is $ \frac{\frac{1}{8}A}{\frac{1}{2}A}=1:4. $