$B_{1}=\frac{\mu_{0} I_{1}}{2 r_{1}} ;$ and $B_{2}=\frac{\mu_{0} I_{2}}{2 r_{2}}$

Since $B_{1}=B_{2} ;$ and $r_{1}=2 r_{2}$

$I_{1}=2 I_{2}$

Again if $R_{1}$ and $R_{2}$ are resistance of the coil $1$ and $2$ then $R_{1}=2 R_{2}($ as $R \propto$ length $=2 \pi r)$ and if $V_{1}$ and $V_{2}$ are the potential difference across them respectively, then

$\frac{V_{1}}{V_{2}}=\frac{I_{1} R_{1}}{I_{2} R_{2}}=\frac{\left(2 I_{2}\right)\left(2 R_{2}\right)}{I_{2} R_{2}}=4$