Answer
The angle between the slides is $0.012^o$.
Work Step by Step
The interference is purely due to the wedge-shaped air gas between the two slides. For a dark fringe to be observed from the top, the rays reflecting off the top edge and the bottom edge of the air-wedge must be exactly out of phase. Since the ray reflecting off the bottom edge of the wedge flips its phase by 0.5\lambda, the additional phase shift due to the path difference must introduce an additional phase shift according to:
$2d_1=m_1\lambda$
Where $d_1$ is the height of the wedge at the point of incidence of the light ray.
Considering the very next dark fringe where the wedge height is $d_2$, we again expect:
$2d_2=(m_1+1)\lambda$
Subtracting the above two equations,
$d_2-d_1={\lambda\over 2}$
From geometry, we see that the angle between the slides is given by:
\begin{align*}
\tan\theta&={d_2-d_1\over x}\\
&={250\times 10^{-9}\over 1.2\times 10^{-3}}\\
&=208.3\times 10^{-6}\\
\theta&=\arctan{208.3\times 10^{-6}}\\
\theta&=0.012^o
\end{align*}
Therefore, the angle between the slides is $0.012^o$.
