In Young's double-slit experiment, the fringe width ($\beta$) can be calculated using the formula:

$$ \beta = \frac{\lambda' \cdot D}{d} $$

where:

$\lambda'$ is the wavelength of light in the medium,

$D$ is the distance between the slits and the screen,

$d$ is the separation between the slits.

The wavelength in the medium ($\lambda'$) is given by:

$$ \lambda' = \frac{\lambda_0}{n} $$

where:

$\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air,

$n = 1.44$ is the refractive index of the liquid.

Substituting the values,

$$ \lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm} $$

Now, convert $\lambda'$ to meters:

$$ \lambda' = 479.17 \, \text{nm} = 479.17 \times 10^{-9} \, \text{m} $$

Next, use the values for $D$ and $d$:

$D = 0.72 \, \text{m}$

$d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$

Calculating the fringe width:

$$ \beta = \frac{479.17 \times 10^{-9} \, \text{m} \cdot 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $$

$$ \beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} $$

$$ \beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm} $$

The closest answer choice is:

Option D: 0.23 mm