An oil drop of mass m carrying a charge +q is held stationary in air by an electric field E. The electric field is correctly given by

$E=mg/q$ acting upwards

The correct answer is Option (1) → $E=mg/q$ acting upwards

For an oil drop of mass $m$ and charge $+q$ held stationary in an electric field $E$:

Force equilibrium: Electric force = Weight

$q E = m g$

Therefore, the required electric field:

$E = \frac{m g}{q}$ Acting Upwards