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

The magnetic field due to an infinitely 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 point $(-1, 1)$ is at a perpendicular distance $d_1 = 1 \text{ m}$ from the $x$-axis. Using the right-hand thumb rule, the magnetic field $\vec{B}_1$ at this point is directed along the positive $z$-axis ($+\hat{k}$).

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

For the wire along the $y$-axis, the current is in the $+\hat{j}$ direction. The point $(-1, 1)$ is at a perpendicular distance $d_2 = |-1| = 1 \text{ m}$ from the $y$-axis. Using the right-hand thumb rule, the magnetic field $\vec{B}_2$ at this point is also directed along the positive $z$-axis ($+\hat{k}$).

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

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

$\vec{B} = \vec{B}_1 + \vec{B}_2 = 1 \mu\text{T} \hat{k} + 1 \mu\text{T} \hat{k} = 2 \mu\text{T} \hat{k}$

Thus, the net magnetic field is $2 \mu\text{T}$ along the positive $z$-axis.