Upon a six fold increase in the external resistance of a circuit, the voltage across the terminals of the battery has increased from 5V to 10V. The  e.m.f. of the battery

12.5V

$ E \frac{R}{R+r} = 5V$   ...........(1)

$ E \frac{6R}{6R+r} = 10 V$   ...........(2) 

Dividing equation (1) and (2)

$\frac{R}{6R} \frac{6R+r}{R+r} = \frac{5}{10}$

$\Rightarrow \frac{6R+r}{R+r} = 3 $

$ 6R+r = 3R + 3r$

$ r = 1.5R$

$\Rightarrow E = 5\frac{R+r}{R} = 12.5V$