Let number of turns $= n$
$n \times 6 a =24 a$
$n =4$
Now magnetic moment $M = nIA =4 . I . A$
Now area of hexagon $\frac{1}{2} a^{2} \sin (120)+a 2 a \sin 60+\frac{1}{2} a^{2} \sin (120)$
$=\frac{\sqrt{3} a ^{2}}{4}+\sqrt{3} a ^{2}+\frac{\sqrt{3} a ^{2}}{4}=\frac{3 \sqrt{3} a ^{2}}{2}$
So magnetic moment $=4$.I. $\left(\frac{3 \sqrt{3} a ^{2}}{2}\right)=6 \sqrt{3} Ia ^{2}$
