The horizontal and vertical components of earth's magnetic field ($B_H$ and $B_V$) are related as

$\frac{B_V}{B_H}=\tan \theta$

Here, $\theta =45°$ and $B_H=18\times {10}^{-6} \mathrm{T}$

$\Rightarrow B_V=B_H\tan 45°$

$\Rightarrow B_V=B_H=18\times {10}^{-6} \mathrm{T}$   $\left(\because \tan 45°=1\right)$

Now, when the external force $F$ is applied, to keep the needle stays in horizontal position is shown below,

Taking torque at point $P$, we get

$m{B}_V\times 2l=Fl$

$\therefore F=2\times m{B}_V$

Substituting the given values, we get

$F=2\times 1.8\times 18\times {10}^{-6}$

$F=6.48\times {10}^{-5}=6.5\times {10}^{-5} \mathrm{N}$.