Answer
$\sqrt{2}$. We cannot use the small-angle approximation.
Work Step by Step
We are told that the distance between the first dark fringes on either side of the central maximum is equal to twice the distance of the screen from the slit. Therefore the distance between the centerline and the first dark fringe equals the distance of the screen from the slit. Call this distance x.
Let $\theta_1$ be the angular position of the first dark fringe. $tan \theta_1=\frac{x}{x}=1$, so the angle $\theta_1=45^{\circ}$.
For the first dark fringe, we have $a sin \theta_1=\lambda$, so $\frac{a}{\lambda}=1/sin \theta_1=1/sin 45^{\circ}$
$$\frac{a}{\lambda}=\sqrt 2$$
