Four long, straight wires, each carrying a current of 5.0 A , are...

Let the centre of the square be the origin $(0,0)$. The side length of the square is $a = 5.0 \text{ cm} = 0.05 \text{ m}$.

The four wires are located at $x = -a/2$, $x = a/2$, $y = -a/2$, and $y = a/2$. The currents are $I = 5.0 \text{ A}$ in the $+y$ direction for the vertical wires and in the $+x$ direction for the horizontal wires.

The diagonal of the square is $d = a\sqrt{2}$. The points $Q_1, Q_2, Q_3, Q_4$ are at a distance $d$ from the origin along the diagonals. Their coordinates are:
$Q_1 = (-a, a)$
$Q_2 = (a, a)$
$Q_3 = (a, -a)$
$Q_4 = (-a, -a)$

The magnetic field due to a long straight wire is $B = \dfrac{\mu_0 I}{2\pi r}$.

At $Q_1(-a, a)$:
Field due to wire at $y = a/2$: $B_1 = \dfrac{\mu_0 I}{2\pi (a - a/2)} = \dfrac{\mu_0 I}{\pi a}$ ($\odot$)
Field due to wire at $y = -a/2$: $B_2 = \dfrac{\mu_0 I}{2\pi (a - (-a/2))} = \dfrac{\mu_0 I}{3\pi a}$ ($\odot$)
Field due to wire at $x = -a/2$: $B_3 = \dfrac{\mu_0 I}{2\pi |-a - (-a/2)|} = \dfrac{\mu_0 I}{\pi a}$ ($\odot$)
Field due to wire at $x = a/2$: $B_4 = \dfrac{\mu_0 I}{2\pi |-a - a/2|} = \dfrac{\mu_0 I}{3\pi a}$ ($\odot$)
Total field at $Q_1$: $B_{Q1} = B_1 + B_2 + B_3 + B_4 = \dfrac{8\mu_0 I}{3\pi a}$ ($\odot$)
Substituting the values: $B_{Q1} = \dfrac{8 \times 4\pi \times 10^{-7} \times 5}{3\pi \times 0.05} = 1.067 \times 10^{-4} \text{ T} \approx 1.1 \times 10^{-4} \text{ T}$ ($\odot$)

At $Q_2(a, a)$:
Fields from horizontal wires are $\odot$ with magnitudes $\dfrac{\mu_0 I}{\pi a}$ and $\dfrac{\mu_0 I}{3\pi a}$.
Fields from vertical wires are $\otimes$ with magnitudes $\dfrac{\mu_0 I}{3\pi a}$ and $\dfrac{\mu_0 I}{\pi a}$.
Total field at $Q_2$: $B_{Q2} = \dfrac{\mu_0 I}{\pi a} + \dfrac{\mu_0 I}{3\pi a} - \dfrac{\mu_0 I}{3\pi a} - \dfrac{\mu_0 I}{\pi a} = 0$

At $Q_3(a, -a)$:
All four wires produce a magnetic field into the page ($\otimes$).
Total field at $Q_3$: $B_{Q3} = \dfrac{8\mu_0 I}{3\pi a}$ ($\otimes$) $\approx 1.1 \times 10^{-4} \text{ T}$ ($\otimes$)

At $Q_4(-a, -a)$:
Fields from horizontal wires are $\otimes$ with magnitudes $\dfrac{\mu_0 I}{3\pi a}$ and $\dfrac{\mu_0 I}{\pi a}$.
Fields from vertical wires are $\odot$ with magnitudes $\dfrac{\mu_0 I}{\pi a}$ and $\dfrac{\mu_0 I}{3\pi a}$.
Total field at $Q_4$: $B_{Q4} = -\dfrac{\mu_0 I}{3\pi a} - \dfrac{\mu_0 I}{\pi a} + \dfrac{\mu_0 I}{\pi a} + \dfrac{\mu_0 I}{3\pi a} = 0$

Answer: $Q_1$: $1.1 \times 10^{-4} \text{ T}$, $\odot$; $Q_2$: zero; $Q_3$: $1.1 \times 10^{-4} \text{ T}$, $\otimes$; $Q_4$: zero