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 carrying current in the positive $x$-direction, the magnetic field at $(1 \text{ m}, 1 \text{ m})$ is directed along the positive $z$-axis (using the right-hand thumb rule).

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

For the wire along the $y$-axis carrying current in the positive $y$-direction, the magnetic field at $(1 \text{ m}, 1 \text{ m})$ is directed along the negative $z$-axis.

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

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

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