The answer is 1.8 * 10^(-7)1.8⋅10−7 "M"M.
The general reaction looks like this
Mg_((aq))^(2+) + 2NaOH_((aq)) -> Mg(OH)_(2(s)) + 2Na^(+)(aq)Mg2+(aq)+2NaOH(aq)→Mg(OH)2(s)+2Na+(aq)
The reaction that is of interest is actually this
Mg(OH)2(s) rightleftharpoons Mg_((aq))^(2+) + 2OH_((aq))^(-)Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)
In order to determine the maximum concentration of Mg^(2+)Mg2+ ions permissible in the NaOHNaOH solution before a precipitate will be formed, you'd need the value of the solubility product constant, K_(sp)Ksp.
The expression for K_(sp)Ksp for this reaction is
K_(sp) = [Mg^(2+)] * [OH^(-)]^2Ksp=[Mg2+]⋅[OH−]2
A precipitate will form when the solution is saturated with ions; if the concentrations of the ions are low enough, a precipitate will not form. The threshold for the formation of a precipitate implies that the above equation be satisfied.
For a saturated solution, on the verge of forming a precipitate, the maximum value for the product of the concentrations of the two ions is equal to K_(sp)Ksp.
From this point on, any increase in the concentrations of the ions will result in the formation of the precipitate.
Since your reaction takes place in 1 L, 0.01 M NaOHNaOH solution, and knowing that NaOHNaOH dissociates completely into Na^(+)Na+ and OH^(-)OH− ions, the concentration of hydroxide ions will be
[OH^(-)] = [NaOH] = 0.01[OH−]=[NaOH]=0.01 "M"M
This means that the maximum concentration of Mg^(2+)Mg2+ ions the solution can hold before a precipitate is formed is
[Mg^(2+)] = K_(sp)/([OH^(-)]^2) = K_(sp)/(0.01)^2 = K_(sp)/(10^(-4)[Mg2+]=Ksp[OH−]2=Ksp(0.01)2=Ksp10−4
As a numerical value, I've found K_(sp)Ksp to be equal to 1.8*10^(-11)1.8⋅10−11 (you can use any value given to you, of course), which means that
[Mg^(2+)] = (1.8 * 10^(-11))/10^(-4) = 1.8 * 10^(-7)[Mg2+]=1.8⋅10−1110−4=1.8⋅10−7 "M"M