Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that the wire is vertical and the point is in the $x$-directino, to the right from our wire. SEE THE FIRST FIGURE BELOW.
Each segment of the wire generates a magnetic field of $dB$ which is given by
$$dB=\dfrac{\mu_0 I (dy)\sin\theta}{4\pi r^2 }$$
where $\sin\theta=d/r$, so
$$dB=\dfrac{\mu_0 I (dy)d}{4\pi r^3 }$$
where $r =\sqrt{(y^2+d^2)}$, and hence
$$dB=\dfrac{\mu_0 I d(dy)}{4\pi (y^2+d^2)^{3/2} }$$
Integrating both sides,
$$\int_0^B dB=\int_{-L/2}^{L/2}\dfrac{\mu_0 I d (dy)}{4\pi (y^2+d^2)^{3/2} }$$
$$B=\dfrac{\mu_0Id}{4\pi}\int_{-L/2}^{L/2} (y^2+d^2)^{-3/2} dy $$
$$B=\dfrac{\mu_0Id}{4\pi}\dfrac{y}{d^2\sqrt{y^2+d^2}}\bigg|_{-L/2}^{L/2} $$
$$B=\dfrac{\mu_0I }{4\pi d}\left[\dfrac{(L/2)}{ \sqrt{(L/2)^2+d^2}} -\dfrac{(-L/2)}{ \sqrt{(-L/2)^2+d^2}}\right]$$
$$\boxed{B=\dfrac{\mu_0I }{4\pi d}\left[\dfrac{L}{ \sqrt{(L/2)^2+d^2}} \right]}$$
$$\color{blue}{\bf [b]}$$
We have here 4 wires which are forming a loop, as seen below.
SEE THE SECOND FIGURE BELOW.
The direction of the magnetic field at the center of this loop is into the page from each wire, so the net magnetic field is the sum of 4 fields.
We can use the boxed formula above, where $L=2R$, and hence $d=R$
$$ B_{\rm net}=4B=\dfrac{4\mu_0I }{4\pi R}\left[\dfrac{(2R)}{ \sqrt{(2R/2)^2+R^2}} \right] $$
$$ B_{\rm net} =\dfrac{\mu_0I }{ \pi R}\left[\dfrac{ 2}{\sqrt2} \right] $$
$$\boxed{ B_{\rm square} =\dfrac{\sqrt{2}\mu_0I }{ \pi R} }$$
$$\color{blue}{\bf [c]}$$
$$\dfrac{B_{\rm square}}{B_{\rm circle}}=\dfrac{\dfrac{\sqrt{2}\mu_0I }{ \pi R} }{\dfrac{ \mu_0I }{ 2 R} }=\dfrac{2\sqrt2}{\pi}$$
$$\dfrac{B_{\rm square}}{B_{\rm circle}}=\color{red}{\bf 0.900} $$
The magnetic field of the square loop equals 0.9 that of the circular loop.
