Solution :
$l_1= 76.3 \;cm$
$l_2 = 64.8\;cm$
$R= 9.5 \; \Omega$
$r= \bigg(\large\frac{l_1}{l_2}$$-1 \bigg)$
Substituting the values
$r= \bigg( \large\frac{76.3}{64.8} $$-1\bigg) \times 9.5$
$\qquad= 1.68 \Omega$
Hence the internal resistance of the cell is $1.68 \Omega$
Answer : $1.68\;\Omega$