$$V_z = 6 V$$
$$V_{max} = 10 V$$
In zener diode
$$I_s = I_{knee} + I_R$$
where $$I_s = $$ supply current by $$10 V$$
$$= \dfrac{V_{max} - V_z}{R}$$
$$I_R = \dfrac{V_{max} - V_z}{R} - I_{knee}$$
$$= \dfrac{10 - 6}{50} - \dfrac{5}{1000} = \dfrac{75}{1000} A$$.
$$R = \dfrac{V_z}{I_R} = \dfrac{6}{\dfrac{75}{1000}} = \dfrac{6 \times 1000}{75} = 80 \Omega$$