Two identical long current carrying wires are bent into the shapes...

For the first figure, the magnetic field at the centre P is the vector sum of the magnetic fields due to two semi-infinite straight wires and one semicircular arc.

The magnetic field due to the top semi-infinite wire at P is $B_{1a} = \dfrac{\mu_0 I}{4\pi r}$ directed into the plane.

The magnetic field due to the bottom semi-infinite wire at P is $B_{1b} = \dfrac{\mu_0 I}{4\pi r}$ directed into the plane.

The magnetic field due to the semicircular arc at P is $B_{1c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}$ directed into the plane.

The total magnetic field at P is $B_1 = B_{1a} + B_{1b} + B_{1c} = \dfrac{\mu_0 I}{4\pi r} (1 + 1 + \pi) = \dfrac{\mu_0 I}{4\pi r} (2 + \pi)$.

For the second figure, the magnetic field at the centre Q is the vector sum of the magnetic fields due to one horizontal semi-infinite wire, one semicircular arc, and one vertical semi-infinite wire.

The magnetic field due to the horizontal semi-infinite wire at Q is $B_{2a} = \dfrac{\mu_0 I}{4\pi r}$ directed into the plane.

The magnetic field due to the semicircular arc at Q is $B_{2c} = \dfrac{\mu_0 I}{4r} = \dfrac{\pi \mu_0 I}{4\pi r}$ directed into the plane.

Since the arc is semicircular, it ends at a point directly below Q. The vertical semi-infinite wire starts from this point and extends downwards, meaning its line of action passes through Q. Therefore, the magnetic field due to this vertical wire at Q is $B_{2b} = 0$.

The total magnetic field at Q is $B_2 = B_{2a} + B_{2b} + B_{2c} = \dfrac{\mu_0 I}{4\pi r} (1 + 0 + \pi) = \dfrac{\mu_0 I}{4\pi r} (1 + \pi)$.

The ratio of the magnetic fields is $\dfrac{B_1}{B_2} = \dfrac{\dfrac{\mu_0 I}{4\pi r} (2 + \pi)}{\dfrac{\mu_0 I}{4\pi r} (1 + \pi)} = \dfrac{2 + \pi}{1 + \pi}$.

Answer: $\dfrac{2 + \pi}{1 + \pi}$