Answer
1.24.
Work Step by Step
The image and object distances are related to the indices of refraction and to the radius of curvature.
$$\frac{n_a}{s}+\frac{n_b}{s’}=\frac{n_b-n_a}{R}$$
$$\frac{n_a}{14.0cm}+\frac{1.60}{-9.00cm}=\frac{1.60-n_a}{-4.00cm}$$
$$0.071n_a-0.1778=-0.4+0.25n_a$$
$$n_a=1.24$$
This is a reasonable refractive index for a liquid.
