When apparatus is immersed in liquid of refractive index $\mu $, fringe width,
$\beta ' = \frac{\beta}{\mu}$
Now $\beta' = \frac{75}{100} \beta$
$ \therefore \, \, \frac{75}{100} \beta = \frac{\beta }{\mu} or \mu = \frac{100}{75} = \frac{4}{3} = 1.33 $