We know that, fring width

$\beta =\frac{D\lambda }{d}$ ....(i)

When distance slits is increased 10 times and distance between between the source and screen is halved, then new fringe width,

${\beta }^{\prime }=\frac{D^{\prime }\lambda }{d\prime }$

$=\frac{D\lambda }{2\times 10\times d}$

$=\frac{1}{20}\cdot \frac{D\lambda }{d}$

$=\frac{\beta }{20}$ [from Equation (i)]