Answer
The liquid's index of refraction is $~~1.46$
Work Step by Step
We can find the angle $\theta_1$ from the normal to the canister's actual position in the pool:
$tan~\theta_1 = \frac{21~cm}{150~cm}$
$\theta_1 = tan^{-1}~(\frac{21~cm}{150~cm})$
$\theta_1 = 7.97^{\circ}$
We can find the angle $\theta_2$ from the normal to the canister's apparent position if we look from the edge of the pool:
$tan~\theta_2 = \frac{31~cm}{150~cm}$
$\theta_2 = tan^{-1}~(\frac{31~cm}{150~cm})$
$\theta_2 = 11.68^{\circ}$
We can find $n_1$, the liquid's index of refraction:
$n_1~sin~\theta_1 = n_2~sin~\theta_2$
$n_1 = \frac{n_2~sin~\theta_2}{sin~\theta_1}$
$n_1 = \frac{1.00~sin~11.68^{\circ}}{sin~7.97^{\circ}}$
$n_1 = 1.46$
The liquid's index of refraction is $~~1.46$
