Answer
The angular deviation is $~~12.4^{\circ}$
Work Step by Step
We can find the angular deviation of the third-order bright fringe:
$\theta_m = sin^{-1}~(\frac{m\lambda}{d})$
$\theta_3 = sin^{-1}~[\frac{(3)(550\times 10^{-9}~nm)}{7.70\times 10^{-6}~m}]$
$\theta_3 = 12.4^{\circ}$
The angular deviation is $~~12.4^{\circ}$
