Answer
$55\,\mu T$
Work Step by Step
Given, angle of dip $\delta=73^{\circ}$
Horizontal component of the earth's magnetic field $H_{E}=16\,\mu T$
Let the magnitude of the earth's magnetic field be $B_{E}$.
Using the formula $H_{E}=B_{E}\cos\delta$, we have
$B_{E}=\frac{H_{E}}{\cos\delta}=\frac{16\,\mu T}{\cos 73^{\circ}}=55\,\mu T$
