Answer
$633\;\rm nm$
Work Step by Step
In single-slit diffraction, we know that the dark fringes (the minima) are given by
$$a\sin\theta_p=p\lambda$$
And for no minima on the screen means that the central maximum is covering the whole screen and the first-order minima is at an angle of $\theta_1=90^\circ$.
Hence,
$$a\sin90^\circ= \lambda$$
Therefore,
$$ a=\lambda=\color{red}{\bf 633}\;\rm nm $$
So the width of the slit must be equal to the wavelength of the light.
