Number of fringes shifted,
$N= \frac{shift}{fringe width} =\frac{\left(\mu -1\right)t}{\lambda}$
or, $ N \propto\left(\mu -1\right)t $ (for same $ \lambda$)
$\therefore \, \frac{N_{1}}{N_{2}} = \frac{\left(\mu_{1} - 1 \right)t_{1}}{\left(\mu_{2} - 1\right)t_{2}}$
Here, $\mu_{1} = 1.5 , t_{1} =t, N_{1} =20 \mu_{2} =1.6, t_{2} = t 2 , N_{2} =? $
or, $\frac{20}{N_{2} } = \frac{\left(1.5 -1\right)t}{\left(1.6 -1\right)\left(t 2\right)} = \frac{2 \times0.5 }{0.6}$
or, $ N_{2} =12$