Two long, straight wires are positioned along the x - and y -axes,...

The magnetic field due to a long straight wire carrying current $I$ at a perpendicular distance $r$ is given by $B = \dfrac{\mu_0 I}{2\pi r}$.

For the wire along the $x$-axis, the current $I_1 = 5 \text{ A}$ flows in the $+\hat{i}$ direction. The point $(1 \text{ m}, -1 \text{ m})$ is at a perpendicular distance $r_1 = 1 \text{ m}$ from the $x$-axis.

Using the right-hand rule, the magnetic field $\vec{B}_1$ at this point is directed into the plane (negative $z$-axis).

$\vec{B}_1 = \dfrac{\mu_0 I_1}{2\pi r_1} (-\hat{k}) = \dfrac{4\pi \times 10^{-7} \times 5}{2\pi \times 1} (-\hat{k}) = -1 \mu\text{T} \hat{k}$

For the wire along the $y$-axis, the current $I_2 = 5 \text{ A}$ flows in the $+\hat{j}$ direction. The point $(1 \text{ m}, -1 \text{ m})$ is at a perpendicular distance $r_2 = 1 \text{ m}$ from the $y$-axis.

Using the right-hand rule, the magnetic field $\vec{B}_2$ at this point is also directed into the plane (negative $z$-axis).

$\vec{B}_2 = \dfrac{\mu_0 I_2}{2\pi r_2} (-\hat{k}) = \dfrac{4\pi \times 10^{-7} \times 5}{2\pi \times 1} (-\hat{k}) = -1 \mu\text{T} \hat{k}$

The net magnetic field is $\vec{B} = \vec{B}_1 + \vec{B}_2 = -1 \mu\text{T} \hat{k} - 1 \mu\text{T} \hat{k} = -2 \mu\text{T} \hat{k}$.

The magnetic field is $2 \mu\text{T}$ along the negative $z$-axis.