(a) The ray in medium B is moving towards normal when it enters from medium $\mathit{A}$. Hence, medium B is optically denser than medium $\mathrm{A}$. This means that the refractive index of medium $\mathit{B}$ with respect to refractive index of medium A will be greater than 1.
(b) Speed of light in rarer medium $\mathrm{A}\left({\mathit{v}}_{\mathit{a}}\right)$ is greater than the speed of light in denser medium $\mathrm{B}\left({\mathit{v}}_{\mathit{b}}\right)$ i.e., ${\mathit{v}}_{\mathit{a}}>{\mathit{v}}_{\mathit{b}}$
Thus, refractive index of $\mathit{B}$ relative to $\mathit{A}$,
$\mathit{n}={\mathit{v}}_{\mathit{a}}∕{\mathit{v}}_{\mathit{b}}$
$\Rightarrow \mathit{n}>1$
OR
(b) Refractive index of water w.r.t. air is $=1.33$ Similarly refractive index of diamond w.r.t. air is $=2.4$
So, Refractive index of water with respect to diamond is $=\frac{1.33}{2.4}=0.55$