Step 1
We are given with the following charge arrangement:



According to Coulomb's law the magnitude of the electric force between two stationary charges is directly proportional to the magnitude of the product of the two charges and inversely proportional to square of the distance between them. Hence the formula for electric field between two charges is: \[F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}\]Here, $q_1$ and $q_2$ are charges on two particles. $r$ is the distance between them and k is constant of permittivity and has a value of:\[K = 9 \times {10^9}\,\,\,{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\]

Step 2: Free Body Diagram of Forces
The charges on particle-2 and particle-3 induce repulsive (due to similar charge) on Particle-1. The forces act in the direction of line joining the respective particles. The free body diagram of the forces is shown below: $F_{12}$ is the force on particle-1 due to particle-2 and $F_{13}$ is the force on particle-1 due to particle-3.

Step 3: The force vectors
Using the coulomb's law: \[\begin{array}{l}{F_{12}} = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}\\ = \dfrac{{9 \times {{10}^9} \times \left( {1 \times {{10}^{ - 9}}\,\,C} \right) \times \left( {2 \times {{10}^{ - 9}}\,\,C} \right)}}{{{{\left( {{{10}^{ - 2}}\,\,{\rm{m}}} \right)}^2}}}\\ = 1.8 \times {10^{ - 4}}\,\,{\rm{N}}\end{array}\]\[\begin{array}{l}{F_{13}} = \dfrac{{K{q_1}{q_3}}}{{{r^2}}}\\ = \dfrac{{9 \times {{10}^9} \times \left( {1 \times {{10}^{ - 9}}\,\,C} \right) \times \left( { 2 \times {{10}^{ - 9}}\,\,C} \right)}}{{{{\left( {{{10}^{ - 2}}\,\,{\rm{m}}} \right)}^2}}}\\ = 1.8 \times {10^{ - 4}}\,\,{\rm{N}}\end{array}\]

Step 4: The magnitude of resultant force
\[\begin{array}{*{20}{l}}{F = \sqrt {F_{12}^2 + F_{13}^2 + 2{F_{12}}{F_{13}}\cos \theta } }\\{ = \sqrt {{{\left( {1.8 \times {{10}^{ - 4}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}} \right)}^2} + {{\left( {1.8 \times {{10}^{ - 4}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}} \right)}^2} + 2\left( {1.8 \times {{10}^{ - 4}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}} \right)\left( {1.8 \times {{10}^{ - 4}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}} \right)\cos 60^\circ } }\\{ = {{10}^{ - 4}}\sqrt {3.24 + 3.24 + 2 \times 3.24\left( {0.5} \right)} }\\{ = 3.11 \times {{10}^{ - 4}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}}\end{array}\]

Step 5: The direction of resultant force
\[\begin{array}{*{20}{l}}{\alpha = {{\tan }^{ - 1}}\left( {\dfrac{{{F_{12}}\sin \theta }}{{{F_{13}} + {F_{12}}\cos \theta }}} \right)}\\{ = {{\tan }^{ - 1}}\left( {\dfrac{{1.8\sin \left( {60^\circ } \right)}}{{1.8 + 1.8\cos \left( {60^\circ } \right)}}} \right)}\\{ = {{\tan }^{ - 1}}\left( {\dfrac{{1.5584}}{{2.7}}} \right)}\\{ = {{\tan }^{ - 1}}\left( {0.5771851} \right)}\\{ = 30^\circ }\end{array}\]So the resultant makes an angle of $30^\circ $ with either force hence it is in the y direction.

ANSWER
Resultant is

$F = 3.11 \times {10^{ - 4}}\,\,{\rm{N}}$ in positive y direction