Answer
$ \dfrac{2}{3} $
Work Step by Step
We know that the intensity of the double-slit interference pattern at position $y$ is given by
$$I=4I_1\cos^2\left(\dfrac{\pi d y}{\lambda L}\right)$$
where we need to find the position at which $I=I_1$
$$\color{red}{\bf\not} I_1=4\color{red}{\bf\not} I_1\cos^2\left(\dfrac{\pi d y}{\lambda L}\right)$$
$$\cos^2\left(\dfrac{\pi d y}{\lambda L}\right)=\dfrac{1}{4}$$
Taking the square root for both sides;
$$\cos\left(\dfrac{\pi d y}{\lambda L}\right)=\dfrac{1}{2}$$
Hence,
$$\dfrac{\pi d y}{\lambda L}=\cos^{-1}\left( \dfrac{1}{2}\right)$$
So,
$$y=\dfrac{\lambda L}{\pi d }\cos^{-1}\left( \dfrac{1}{2}\right)\tag 1$$
Now we need to find the distance to the first minimum $y_0$, where the dark fringe in a double-slit experiment is given by
$$y_m=\left(m+\frac{1}{2}\right)\dfrac{\lambda L}{d}$$
The first minimum is at $m=0$
$$y_0=\left(0+\frac{1}{2}\right)\dfrac{\lambda L}{d}$$
Hence,
$$y_0=\dfrac{\lambda L}{2d}\tag2$$
To find the needed ratio, divide (1) by (2);
$$\dfrac{y}{y_0}=\dfrac{\dfrac{\lambda L}{\pi d }\cos^{-1}\left( \dfrac{1}{2}\right)}{\dfrac{\lambda L}{2d}}=\dfrac{2}{\pi}\;\;\overbrace{\cos^{-1}\left( \dfrac{1}{2}\right)}^{\pi/3}$$
$$\dfrac{y}{y_0}=\color{red}{\bf \dfrac{2}{3} }$$
