From the interference formula: $sin\theta =\frac{m\lambda }{d }$ , considering that at third maximum m = 3, we obtain:

$\lambda =\frac{d\cdot sin\theta }{m}=\frac{0.0334\cdot 10^{-3}sin\left(3.21°\right)}{3}=\mathrm{6.23..}\cdot 10^{-7}m\simeq 623nm$