Let intensity of light on screen due to each slit is $I_0$.

So internity at centre of screen is $4I_0$(as $\cos \varphi =1$). 

Intensity at distance $y$ from centre, 

$I=I_0+I_0+2\sqrt{I_0{I}_0}\cos \varphi$

As we know, $I_{\max }=4I_0$. Therefore,

$\Rightarrow \frac{I_{\max }}{2}=2I_0=2I_0+2I_0\cos \varphi$

$\Rightarrow \cos \varphi =0$

$\Rightarrow \varphi =\frac{\pi }{2}$

Hence, $k\Delta \mathrm{x}=\frac{\pi }{2}$

$\Rightarrow \frac{2\pi }{\lambda }\left(\mathrm{dsin}\theta \right)=\frac{\pi }{2}$

$\Rightarrow \frac{2}{\lambda }d\times \frac{y}{D}=\frac{1}{2}$

$\Rightarrow y=\frac{\lambda D}{4 d}=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}$

$=125\times {10}^{-6}$

$=125$