Solution :
given : $G= 10 \Omega$
$I_g= 1mA$
$\quad= 10^{-3} \;A$
$V=2$
$R_1= \large\frac{V}{I_g}$$-G$
$\quad= \large\frac{2}{10^{-3}}$$-10$
$\quad= 1990 \Omega$
$\qquad = 2k \Omega$
(ii) $v= 20\;v ,(R_1+R_2) =\large\frac{20}{10^{-3}}$$-10=20,000 -10^{\circ}$
$\quad \approx 20 k \Omega$
(iii) $v= 200 \;v, R_1 +R_2+R_3=\large\frac{200}{10^{-3}}$$-10 \approx 200 \;K \Omega$
$R_1 = 2k \Omega , R_2 = 20 \;k \Omega$
$R_3 = 200 -20 =180 k \Omega $
Answer : $180\;K\;\Omega $