Explanation

(c) Given that
$emf\ E_{N} = 1.5r_{N}$
Where $r_{N}$ is the internal resistance of $n$th cell.
Total emf $E = E_{1} + E_{2} + E_{3} + \ldots + E_{n}$
$\ \ \ = 1.5\left\lbrack r_{1} + r_{2} + r_{3} + \ldots + r_{n} \right\rbrack$
Total internal resistance
$r = r_{1} + r_{2} + r_{3} + \ldots + r_{n}$
$\therefore Current\ i = \frac{E_{total}}{r_{total}}$
$i = \frac{1.5\left\lbrack r_{1} + r_{2} + r_{3} + \ldots + r_{n} \right\rbrack}{\left\lbrack r_{1} + r_{2} + r_{3} + \ldots + r_{n} \right\rbrack}$
Hence, $i = 1.5A$