#### Answer

$F = 0.033~N$

#### Work Step by Step

We can find the time to reach maximum height.
$t_{up} = \frac{v_y-v_{y0}}{a_y}$
$t_{up} = \frac{0 - (6.0~m/s)~sin(35^{\circ})}{-9.80~m/s^2}$
$t_{up} = 0.351~s$
The total time of the object's flight is double the time it takes to reach maximum height.
$t = 2~t_{up} = 0.702~s$
We can find the horizontal acceleration.
$x = v_{0x}~t+\frac{1}{2}a_x~t^2$
$a_x = \frac{2(x-v_{0x}~t)}{t^2}$
$a_x = \frac{(2)[(2.9~m)-(6.0~m/s)~cos(35^{\circ})(0.702~s)]}{(0.702~s)^2}$
$a_x = -2.23~m/s^2$
We can use the magnitude of the horizontal acceleration to find the magnitude of the electric force $F$.
$F = m~a_x$
$F = (0.015~kg)(2.23~m/s^2)$
$F = 0.033~N$