Answer
$\approx$ 27 fringes per cm
Work Step by Step
We can first calculate the angle that is between the air pocket and the top glass pane:
$$
\theta = sin^{-1}(\frac{0.0740mm}{8.35cm}) \approx8,9\cdot 10^{-4} rad
$$
Now, we can say that each fringe will occur at a distance $x$ from the start of the second glass pane. This will give us a triangle exactly like with the air pocket, but with a bottom side $x$ and the height equal to the thickness of the air pocket at a given point.
Since we have one and exactly only one reflection against a medium where the index of refraction is higher (air - glass), constructive interferences occur (fringes are visible) when
$$
2nd=(m+\frac 1 2) \lambda
$$
For any given fringe, according to trigonometry of the relationship established above
$$
d=xtan \implies 2xtan\theta =(m+\frac 1 2) \lambda \implies x = \frac{(m+\frac 1 2)\lambda}{2tan\theta}
$$
Taking the difference between two fringes next to each other:
$$
\Delta x = x_{2}- x_1 = \frac{(m+\frac 3 2)\lambda}{2tan\theta} - \frac{(m+\frac 1 2)\lambda}{2tan\theta} = \frac{(m+\frac 1 2)\lambda}{2tan\theta}
$$
Finally, the number of fringes per m:
$$
\frac{1}{\Delta x}
$$
but the question wants it in cm:
$$
\frac{\frac{1}{\Delta x}}{100} \approx 27 \text{ fringes per cm}
$$
