Answer
The slit width is approximately $3.4\times 10^{-6}m$.
Work Step by Step
See Figure 24-21. The angle from the central maximum to the first bright maximum is $16^{\circ}$, exactly half the angle given (which was the angle between the first bright maxima on either side of the centerline).
The angle to the first maximum is approximately halfway between the angles to the first and the second minima, or m = 1.5. Apply equation 24–3b with m = 1.5 and find the slit width, D.
$$Dsin \theta=m\lambda$$
$$Dsin \theta_{1st \;max}=(1.5)\lambda$$
$$D=\frac{1.5 (633\times10^{-9}m)}{sin 16^{\circ}}\approx 3.4\times 10^{-6}m$$
