The percentage of hydrolysis of 0.003 M aqueous solution of NaOCN  $(K_{a}forHOCNis3.33\times {10}^{-4}M)$ is :

NaOCN is a salt of weak acid and strong base 

$K_h=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{3.33\times {10}^{-4}}$$=0.300\times {10}^{-10}=3.0\times {10}^{-11}$

Degree of hydrolysis

$h=\sqrt{\frac{K_h}{C}}$

$=\sqrt{\frac{3.0\times {10}^{-11}}{0.003}}=\sqrt{\frac{3.0\times {10}^{-11}}{3.0\times {10}^{-3}}}$=$\sqrt{{10}^{-8}}=1\times {10}^{-4}$

Percentage of hydrolysis = degree of hydrolysis  x 100

$=1\times {10}^{-4}\times {10}^2={10}^{-2}=0.01\%$