Answer
506 nm.
Work Step by Step
Calculate the angular position of the first minimum.
$$tan \theta_1=\frac{0.00135m}{2.00m}$$
$$ \theta_1=6.75\times10^{-4}rad$$
Now relate the slit width to the wavelength and this angle.
$$\lambda=a sin \theta_1=(0.750\times10^{-3}m)sin (6.75\times10^{-4}rad)=506 nm$$
