Solution :
Given $n=4$
$I=IA$
$R= 0.39 \;G$
$\quad= 0.39 \times 10^{-4} \;T$
$\delta =35^{\circ}$
$\theta = 0^{\circ}$
distance from the cable $r= 4\;cm$
$\qquad= 0.04 \;m$
Let the magnetic field due to four long wires at P and B
Magnetic field due to an infinitely long current carrying wire is
$B'= \large\frac{\mu_0}{4 \pi} \frac{2I}{r}$
$B= 4 \times \large\frac{\mu_0}{4 \pi} \times \large\frac{2I}{r}$
$\qquad = \large\frac{4 \times 10^{-7} \times 2 \times 1}{4 \times 10^{-2}}$
$\qquad= 2 \times 10^{-5}\;T$
Also ,horizontal component of earth magnetic field
$H= R \cos \delta$
$\qquad= 0.39 \times 10^{-4} \times \cos 35^{\circ}$
$\qquad= 3.19 \times 10^{-5} T$
Vertical component of earth magnetic field is
$V= R \sin \delta$
$\quad= 0.39 \times 10^{-4} \times \sin 35^{\circ}$
$\quad= 0.39 \times 10^{-4} \times 0.5736$
$\quad= 2.2 \times 10^{-5} \;T$
At the point below the cable is
$H'= H-B$
$H'= 3.19 \times 10^{-5} -2 \times 10^{-5}$
$\qquad= 1.19 \times 10^{-5} \;T$
The resultant magnetic field is
$R= \sqrt{(H_1)^2+(v^2)}$
$\qquad= \sqrt {(1.19 \times 10^{-5})^2+(2.2 \times 10^{-5} )^2}$
$\qquad= 2.5 \times 10^{-5} T$
Hence the resultant magnetic field below the cable is $2.5 \times 10^{-5}\;T$
Answer : $2.5 \times 10^{-5}\;T$