Given : ${}_{\mathit{a}}{\mathit{\mu }}_{\mathit{w}}=\frac{4}{3}$, Shift $=3.0\mathrm{mm}$.
Let the real depth of the coin in water be

$\therefore$ Apparent depth of the coin $=\left(\mathit{x}-3\right)\mathrm{mm}$.
We know,

$\therefore \frac{4}{3}=\frac{\mathit{x}}{\left(\mathit{x}-3\right)}$
$\Rightarrow 4\left(\mathit{x}-3\right)=3\mathit{x}$
$\Rightarrow 4\mathit{x}-12=3\mathit{x}$
$\Rightarrow 4\mathit{x}-3\mathit{x}=12$

$\therefore$ Real depth of the coin in water