(c) When a resistance of 100$\Omega$ is connected in series current,
$i = \frac{V}{100 + R}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots\ldots.(i)$
When a resistance of 1000$\Omega$ is connected in series, the its range double
Current, $i = \frac{2V}{1100 + R}$ ……(ii)
From Eqs. (i) and (ii)
$\frac{V}{100 + R} = \frac{2V}{100 + R}$
$R = 900\Omega$