Explanation

(b) Let the resistances be $R_{1},\ R_{2}$ and $R_{3}$
$\therefore\frac{R_{1}}{R_{2}} = \frac{1}{2} \Longrightarrow R_{1} = k,\ R_{2} = 2k$
In parallel $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{1} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{R_{3}}$
$\frac{1}{R_{3}} = 1 - \frac{1}{k} = \frac{1}{2k}$
$\ \ \ \ \ \ = \frac{2k - 2 - 1}{2k} = \frac{2k - 3}{2k}$
$R_{3} = \frac{2k}{2k - 3}$
If k=1, then $R_{3}$ is found to be negative, which is impossible.
If k=2, then $R_{1} = 2,\ R_{2} = 4\ R_{3} = 4$
$R_{2} = R_{3}$, not satisfying the condition of the question that all resistance are unequal.
$If\ k = 3,\ then\ R_{1} = 3,\ R_{2} = 6$
$\ \ \ \ R_{3} = 2\Omega$
∴ Largest resistance =6$\Omega\ $