Two wires of same material A and B, having lengths l and (2l)/(3)...

The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is given by:

$B = \dfrac{\mu_0 I}{2R}$

Since the wire of length $L$ is bent into a single turn circular loop, the circumference is equal to the length of the wire:

$2\pi R = L \Rightarrow R = \dfrac{L}{2\pi}$

Substituting the value of $R$ in the expression for $B$:

$B = \dfrac{\mu_0 I}{2 \left(\dfrac{L}{2\pi}\right)} = \dfrac{\mu_0 \pi I}{L}$

For wire A:

$L_A = l$ and $I_A = I$

$B_A = \dfrac{\mu_0 \pi I}{l}$

For wire B:

$L_B = \dfrac{2l}{3}$ and $I_B = \dfrac{3I}{4}$

$B_B = \dfrac{\mu_0 \pi \left(\dfrac{3I}{4}\right)}{\dfrac{2l}{3}} = \dfrac{9 \mu_0 \pi I}{8l}$

Taking the ratio of the magnetic fields:

$\dfrac{B_B}{B_A} = \dfrac{\dfrac{9 \mu_0 \pi I}{8l}}{\dfrac{\mu_0 \pi I}{l}} = \dfrac{9}{8}$

Answer: $\dfrac{9}{8}$