As, relative error $=\frac{\text { deviation in measurement }}{\text { true measurement }}$

For student $A$,

$RE =\frac{0.5 \times 10^{-2}}{9.5}=0.0005$

For student $B$ the meter-scale should be used $10$ times, $\therefore RE =\frac{0.1 \times 10^{-2}}{1} \times 10=0.01$

For student $C$,

$1$ foot $=30.48 \,cm$

So, the scale should be used $31$ times.

$\therefore RE =\frac{0.05}{30.48} \times 31=0.05$

Hence, the lowest relative error $(RE)$ in measured value is for student $A$.