$\Rightarrow|\overrightarrow{ d }|= rd \theta$

$\Rightarrow|\overrightarrow{ B }|=\frac{\mu_0 I }{2 \pi I }$

$\Rightarrow \int \overrightarrow{ B } \cdot \overrightarrow{ d \ell}=\int|\overrightarrow{ B }||\overrightarrow{ d }| \cos 0^{\circ}$

$=\int\left(\frac{\mu_0 I }{2 \pi r }\right) \times( rd \theta)$

$=\int_{\theta_1}^{\theta_2} \frac{\mu_0 I }{2 \pi} d \theta=\frac{\mu_0 I }{2 \pi}\left[\theta_2-\left(-\theta_1\right)\right]$

[ $\theta_1$ is anticlockwise hence taken negative]

(image)

$\Rightarrow \tan \theta_1=\frac{a \sqrt{3}}{a}=\sqrt{3}$

$\theta_1=\frac{\pi}{3}$

$\Rightarrow \tan \theta_2=\frac{a}{a}=1$

$\theta_2=\frac{\pi}{4}$

$\Rightarrow \int B \cdot d \ell=\frac{\mu_0 I}{2 \pi}\left[\frac{\pi}{3}+\frac{\pi}{4}\right]$

$=\frac{7 \mu_0 I}{24}$

$\Rightarrow$ Ans. Option $(A)$