Dipole moment of circular loop is $m=I A$

$\mathrm{m}_{1}=\mathrm{I} \cdot \mathrm{A}=\mathrm{I} \pi \mathrm{R}^{2} \quad\{\mathrm{R}=\text { Radius of the loop }\}$

If moment is doubled (keeping current constant) $R$ becomes $\sqrt{2 \mathrm{R}}$

${{\text{m}}_2} = {\text{L}}\pi {(\sqrt 2 {\text{R}})^2} = 2.{\text{I}}\pi {{\text{R}}^2} = 2{{\text{m}}_1}$

$B_{2}=\frac{\mu_{0} I}{2(\sqrt{2} R)}$