Answer
$d = 633~nm$
Work Step by Step
We can write the general condition for a minimum:
$d~sin~\theta_m = m~\lambda$
$d = \frac{m~\lambda}{sin~\theta_m}$
We can find the slit width when $\theta_1 = 90^{\circ}$:
$d = \frac{m~\lambda}{sin~\theta_m}$
$d = \frac{(1)~(\lambda)}{sin~\theta_1}$
$d = \frac{633~nm}{sin~90^{\circ}}$
$d = 633~nm$
