Answer
$F = 2.2\times 10^{-3}~N$
Work Step by Step
The vertical component of the tension $~~T_y = mg$
We an find an expression for the horizontal component of the tension:
$\frac{T_x}{T_y} = tan~\theta$
$T_x = T_y~tan~\theta$
$T_x = mg~tan~\theta$
The force $F$ of the push is equal in magnitude to $T_x$
We can find $F$:
$F = T_x$
$F = mg~tan~\theta$
$F = (3.0\times 10^{-4}~kg)(9.8~m/s^2)~tan~37^{\circ}$
$F = 2.2\times 10^{-3}~N$
