electric field

For the whole system in equilibrium, charge Q must be -ve.
Hence, force of attraction F given by Q on a charge q at one of the vertices is,
F = k(-Q)q/[(2/3)a.sin(60)]^2
where k is a constant
and [a.sin(60)] is the length of a median, the centre of the triangle is at a point (2/3) of the length of the median measured from a vertice.

The force of repulsion f given by a charge q on an adjacent vertice
f = kq^2/a^2
hence, total repulsive force f' given by the two charges at the other two vertices
f' = 2f.cos(30) = 2kq^2.cos(30)/a^2

If the whole system is in equilibrium, f' = F
i.e. 2kq^2.cos(30)/a^2 = k(-Q)q/[(2/3)a.sin(60)]^2
simplifying,
Q = -[2 x 4/9 x cos(30).sin^2(60)]q = -0.577q