How many grams of NaCN

You firstly need the following data :
Molar mass of NaCN = 22.99 + 12.01 + 14.01 = 49.01 g mol-1
Ka of HCN = 4.9 x 10-10 M
Kw of H2O = 1 x 10-14 M2
Hence, Kh of CN- = Kw(H2O)/Ka(HCN) = 2.041 x 10-5 M

CN-(aq) + H2O(aq) = HCN(aq) + OH-(aq) aaa(Kh)
At equilibrium :
pH = 10
pOH = 14 - 10 = 4
[OH-] = 10-pOH = 1 x 10-4 M
Both HCN and OH- are simultaneously formed in the same reaction.
Hence, [HCN] = 1 x 10-4 M
Kh = [HCN] [OH-] / [CN-]
Hence, [CN-] = [HCN] [OH-] / Kh
[CN-] = (1 x 10-4) (1 x 10-4) / (2.041 x 10-5) = 4.9 x 10-4 M

Some of CN- has hydrolysed to give HCN and OH-.
Hence, [CN-]o = (4.9 x 10-4) + (1 x 10-4) = 5.9 x 10-4 M
No. of mole of CN- added = MV = (5.9 x 10-4) x (300/1000) = 1.77 x 10-4 mol
No. of mole of NaCN solid added = 1.77 x 10-4 mol
Mass of NaCN = mol x (molar mass) = (1.77 x 10-4) x 49.01 = 0.006875 g