Answer
$m(HCN) = 5.94g$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that 6 moles of $CN^-$ produce 1 mole of $C_2N_2$, and 1 mole of $HCN$ produces 1 mole of $H_2TeO_3$. Hence, 6 moles of $CN^-$ produce 1 mole of $HCN$.
$n(HCN) = \frac{1}{6}n(CN^-) = \frac{1}{6}n(KCN) = \frac{1}{6}\frac{m(KCN)}{M(KCN)}=\frac{1}{6}\frac{85.77g}{65\frac{g}{mol}}=0.22mol\implies m(HCN) = M(HCN)\times n(HCN) = 27\frac{g}{mol}\times 0.22mol=5.94g$
