As shown in the figure, a long wire is bent at its middle to form a...

Let the bend of the wire be at the origin $(0,0)$.

The wire consists of two semi-infinite segments:
1. A vertical segment along the negative y-axis.
2. A horizontal segment along the negative x-axis.

The given points are $P(0, -d)$, $Q(d, 0)$, $R(0, d)$, and $S(-d, 0)$.

Consider the magnetic field at point $P(0, -d)$:
Point $P$ lies on the vertical segment, so the magnetic field due to this segment at $P$ is zero.
Point $P$ is at a perpendicular distance $d$ from the end of the semi-infinite horizontal segment.
The magnetic field due to a semi-infinite wire at a perpendicular distance $d$ from its end is $B = \dfrac{\mu_0 i}{4\pi d}$.
Thus, the total magnetic field at $P$ is $\dfrac{\mu_0 i}{4\pi d}$.

Similarly, point $Q(d, 0)$ lies on the extended axis of the horizontal segment (zero contribution) and is at a perpendicular distance $d$ from the end of the vertical semi-infinite segment. The field at $Q$ is also $\dfrac{\mu_0 i}{4\pi d}$.

By symmetry, the fields at $R$ and $S$ have the same magnitude.

The common magnitude is $\dfrac{\mu_0 i}{4\pi d}$.

Answer: $\dfrac{\mu_0 i}{4\pi d}$