Step 1: Understand the question.
In Young's experiment the screen is $75$ cm from the slits. The slit separation is doubled, which makes the fringes thinner. We find how far to move the screen to bring the fringe width back to its first value.
Step 2: Recall the fringe width formula.
The fringe width is \[ \beta = \frac{\lambda D}{d}, \] where $D$ is the slit-to-screen distance and $d$ is the slit separation.
Step 3: See the effect of doubling $d$.
If $d$ becomes $2d$, the fringe width becomes $\frac{\lambda D}{2d}$, which is half of the original. The fringes get narrower.
Step 4: Decide how to restore $\beta$.
To keep $\beta$ the same while $d$ doubled, $D$ must also double, because $\beta$ depends on the ratio $\frac{D}{d}$. So the new screen distance is $2\times75 = 150$ cm.
Step 5: Find the distance moved.
The screen moves from $75$ cm to $150$ cm, a change of \[ 150 - 75 = 75\ \text{cm}. \] Since the distance increases, the screen moves away from the slits.
Step 6: State the answer.
The screen should be moved $75$ cm away from the slits. \[ \boxed{75\ \text{cm away from the slits}} \]