$\frac{1}{2} xPD X _5 x =\frac{1}{2} x _{3 x } X 4 x$

$\therefore PD =\frac{12 x }{5}$

$QD =\sqrt{( PQ )^2-( PQ )^2}=\sqrt{9 x ^2-\frac{144 x ^2}{25}}=\frac{9 x }{5}$

$\text { and } DR =5 x -\frac{9 x }{5}=\frac{16 x }{5}$

Magnetic field at $P$ due to current elements $P Q$ and $P R$ is zero as the point $P$ is on the conductor. Therefore, magnetic field at $P$ due to current element $QR$ is

$B=\frac{\mu_0 I }{4 \pi PD }\left(\sin \phi_1+\sin \phi_2\right)$

$B =\frac{\mu_0 I \times 5}{4 \pi \times 12 x }\left(\frac{(9 x / 5)}{3 x }+\frac{(16 x / 5)}{4 x }\right)$

$B =\frac{\mu_0 I }{48 \pi x }\left(\frac{3}{5}+\frac{4}{5}\right)$

$B =\frac{7 \mu_0 I }{48 \pi x } \cdot k =7$