The correct option of this question will be (a).
Solution —
Given $q=-16 \times 10^{-18} C , v=10 \hat{ m s } s ^{-1}$
$\vec{B} =B \hat{j}, \vec{E}=-10^{4} \hat{k} Vm ^{-1}$
$\vec{F}_{e} =q \vec{E}=-16 \times 10^{-18} \times\left(-10^{4} \hat{k}\right)$
=$16 \times 10^{-14} \hat{k} N$
$\vec{F}_{m} =q(\vec{v} \times \vec{B})=-16 \times 10^{-18}(10 \hat{i} \times B \hat{j})$
=$-16 \times 10^{-17} B \hat{k} N$
As the particle continues to move along the same direction
$F_{m}=F_{e}$
$\therefore \quad 16 \times 10^{-17}$
$B$ = $16 \times 10^{-14}$
$\text { or } B=10^{3} Wb m ^{-2}$