In the magnetic meridian at a certain place the horizontal component of earths magnetic field is 32 G and the angle of dip is 60°. The total magnetic field of the earth at this location, is:

0.64 G

The correct answer is Option (2) → 0.64 G

$H = 32 G$

$\delta=60^{\circ}$

$H=R \cos \delta$

$R=\frac{H}{\cos \delta}$

$=\frac{32 \times 2}{1} = 64 G$

If H would be 0.32 G then answer would be 0.64 G.