Answer
$623nm$
Work Step by Step
We can find the required wavelength as follows:
$\lambda=\frac{dsin\theta}{m}$
We plug in the known values to obtain:
$\lambda=\frac{(0.0334\times 10^{-3}m)(sin(3.21^{\circ}))}{3}$
$\lambda=6.23\times 10^{-7}m$
$\lambda=623\times 10^{-9}m$
$\lambda=623nm$
