Hint: Use lens makers formula.

Step 1: Find the length of the liquid.
The focal length of the convex lens, f${}_1$=30 cm
The focal length of the liquid = f${}_2$

Focal length of the system (convex lens+liquid), f = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length:

$\frac{1}{\mathrm{f}}=\frac{1}{{\mathrm{f}}_1}+\frac{1}{{\mathrm{f}}_2}$
$\frac{1}{{\mathrm{f}}_2}=\frac{1}{\mathrm{f}}-\frac{1}{{\mathrm{f}}_1}=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$
$\Rightarrow {\mathrm{f}}_2=-90 \mathrm{cm}$

Step 2: Find the radius of curvature of the lens.
Let the refractive index of the lens be ${\mathrm{\mu }}_1$ and the radius of curvature be R.

Using the lens maker's formula:

$\Rightarrow \frac{1}{{\mathrm{f}}_1}=\left({\mathrm{\mu }}_1-1\right)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right) \Rightarrow \frac{1}{30}=\left(1.5-1\right) \left(\frac{2}{\mathrm{R}}\right) \Rightarrow \mathrm{R}=\frac{30}{0.5\times 2}=30 \mathrm{cm}$

Step 3: Find the refractive index of the liquid.
Let ${\mathrm{\mu }}_2$ be the refractive index of the liquid.

Again using the lens maker's formula:

$\frac{1}{{\mathrm{f}}_2}=\left({\mathrm{\mu }}_2-1\right)\left[-\frac{1}{\mathrm{R}}-\frac{1}{\infty }\right]$
$\frac{-1}{90}=\left({\mathrm{\mu }}_2-1\right)\left[-\frac{1}{30}-0\right]$
${\mathrm{\mu }}_2-1=\frac{1}{3}$
$\Rightarrow {\mathrm{\mu }}_2=\frac{4}{3}=1.33$

Hence, the refractive index of the liquid is 1.33.