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 $d$ is given by $B = \dfrac{\mu_0 I}{2\pi d}$.

For the wire along the $x$-axis, the current is in the $+\hat{i}$ direction. The position vector of the point $(-1, -1)$ with respect to the origin on the $x$-axis is in the $-\hat{j}$ direction (since $y = -1 \text{ m}$).

The magnetic field $\vec{B}_1$ due to this wire is:

$\vec{B}_1 = \dfrac{\mu_0 I}{2\pi d} (\hat{i} \times -\hat{j}) = \dfrac{4\pi \times 10^{-7} \times 5}{2\pi \times 1} (-\hat{k}) = -10^{-6} \hat{k} \text{ T} = -1 \mu\text{T} \hat{k}$

For the wire along the $y$-axis, the current is in the $+\hat{j}$ direction. The position vector of the point with respect to the $y$-axis is in the $-\hat{i}$ direction (since $x = -1 \text{ m}$).

The magnetic field $\vec{B}_2$ due to this wire is:

$\vec{B}_2 = \dfrac{\mu_0 I}{2\pi d} (\hat{j} \times -\hat{i}) = \dfrac{4\pi \times 10^{-7} \times 5}{2\pi \times 1} (\hat{k}) = 10^{-6} \hat{k} \text{ T} = 1 \mu\text{T} \hat{k}$

The net magnetic field at the point $(-1 \text{ m}, -1 \text{ m})$ is the vector sum of the individual fields:

$\vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2 = -1 \mu\text{T} \hat{k} + 1 \mu\text{T} \hat{k} = 0$