Answer:
Option C
Explanation:
net force is at $45^{0}$ from vertical
$\therefore$ qE=mg
or $\frac{qX}{d}=mg$ ($\because E=\frac{X}{d}$)
or $X= \frac{mgd}{q}$
= $\frac{(1.67 \times 10^{-27})(9.8)(10^{-2})}{(1.6 \times 10^{-19})}$
=$1 \times 10^{-9} V$