An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :

Split the question into two parts: $$ \vec{B}$$ at point O by infinitely long wire and by circular loop.

 for straight wire :

$$ \vec{B}_1 = \frac{\mu_0 I}{2\pi r} \hat{i} $$

By the Right-Hand Rule,the magnetic field at O points outward, in the +x direction ($$\hat{i}$$)

for circuar loop :$$ \vec{B}_2 = \frac{\mu_0 I}{2r} (-\hat{i}) = -\frac{\mu_0 I}{2r} \hat{i} $$

According to the Right-Hand Rule for circular loops, the magnetic field at the centre points
inward, into the page, which is the −x direction ($$-\hat{i}$$)

The total magnetic field at the centre $$O$$ is the vector sum of $$\vec{B}_1$$ and $$\vec{B}_2$$:
$$ \vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2 $$
$$ \vec{B}_{\text{net}} = \frac{\mu_0 I}{2\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i} $$

Taking $$\frac{\mu_0 I}{2r}$$ as a common factor:
$$ \vec{B}_{\text{net}} = \frac{\mu_0 I}{2r} \left( \frac{1}{\pi} - 1 \right) \hat{i} $$