Answer
$q_1 = +0.750nC$
Work Step by Step
Since $q_3$ is positive, Force due to charge $q_2 = -3nC\space on \space q_3$ would be in positive x direction.
So to cancel this force $F_{q_1q_3}$ should be in negative x direction, hence $q_2$ must be of positive charge with $F_{q_1q_3} = F_{q_2q_3}$
${kq_1q_3\over (2\times 10^{-2})^2} = {kq_2q_3\over (4\times 10^{-2})^2} $
$q_1 = {q_2\over 4} = -{3nC\over 4} = 0.750nC$
