Double check Answer.

T Sin 20° = (1 X 10^{-2} kg) ---Eq1

T Cos 20° = - F ---Eq2 rearranged.

Dividing Eq1 by Eq 2

Tan 20° = (1 X 10^{-2} kg) / (- F)

(- F) = (1 X 10^{-2} kg) / Tan 20°

F = -(2.7 X 10^{-2} N)

Alternative solution:

Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Cos 20° = ^{W}/_{T}

T = ^{W}/_{Cos 20°}

T = ^{(0.01N)}/_{Cos 20°}

T = (1.1 X 10^{-2} N)

Tan 20° = ^{F}/_{W}

F = ^{Tan 20°}/_{W}

F = ^{(Tan 20°)}/_{(0.01N)}

F = (3.6 X 10^{-3} N)

Double check:

(1.1 X 10^{-2} N)^{2} = (3.6 X 10^{-3} N)^{2} +
(1.0 X 10^{-2} N)^{2}