${\mathit{B}}_{\mathit{H}}={\mathit{B}}_{\mathit{E}}\cos \mathit{\delta }$
${\mathit{B}}_{\mathit{H}}={\mathit{B}}_{\mathit{E}}\cos {60}^{\circ }\Rightarrow {\mathit{B}}_{\mathit{E}}=2{\mathit{B}}_{\mathit{H}}$
At equator $\mathit{\delta }=0^{\circ }$

$\therefore {\mathit{B}}_{\mathit{H}}=2\mathit{B}\cos 0^{\circ }=2\mathit{B}$
[Alternatively, award full one mark, if student doesn't take the value $\left(=2\mathit{B}\right)$ of ${\mathit{B}}_{\mathit{E}}$ while finding the value of horizontal component at equator, and just writes the formula only.]