The length, tension, diameter and density of a wire $B$ are double than the corresponding quantities for another stretched wire $A$. Then.
Medium
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We know,

$v =\sqrt{\frac{T}{\mu}}$

Also, $\mu \propto r^2$

Thus,

$v_A \propto \sqrt{\frac{T}{\rho(r)^2}}$

$v_B \propto \sqrt{\frac{2 T}{2 p(2 r)^2}}$

$v_B=\frac{v_A}{2}$

Also,

$n _{ S }=\frac{1}{2(2 L )} \sqrt{\frac{2 T }{8 \mu}}$

$n _{ S }=\frac{1}{8 L } \sqrt{\frac{ T }{\mu}}$

$n _{ A }=4 n _{ B }$

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