Deflection current =Igmax=nxk=0.005×30 Where, n= Number of divisions =30 and k=0.005amp/ division $\begin{array}{l}
=15 \times 10^{-2}=0.15 \
\mathrm{v}=\mathrm{I}_{\mathrm{g}}[20+\mathrm{R}] \
15=0.15[20+\mathrm{R}] \
100=20+\mathrm{R} \
\mathrm{R}=80 \Omega
\end{array}$