From the interference formula:

$sin\theta =\frac{m\lambda }{d }$

considering that at second maximum m = 2, we obtain

$\lambda =\frac{d\cdot sin\theta }{m}= \frac{480\cdot 10^{-6}sin\left(0.099°\right) }{2}=\mathrm{4.1469..}\cdot 10^{-7}m\simeq 415nm$