Correct option (b) \(4450 \Omega\)
Explanation:
Total initial resistance
$$
\begin{aligned}
&=R_{G}+R_{1}=(50+2950) \Omega=3000 \Omega \\
&\varepsilon=3 \mathrm{~V} \\
&\therefore \text { Current }=\frac{3 \mathrm{~V}}{3000 \Omega}=1 \times 10^{-3} \mathrm{~mA}
\end{aligned}
$$
If the deflection has to be reduced to 20 divisions,
current \(i=1 \mathrm{~mA} \times \frac{2}{3}\),
as the full deflection scale for
\(1 \mathrm{~mA}=30\) divisions.
$$
\begin{aligned}
& 3 \mathrm{~V}=3000 \Omega \times 1 \mathrm{~mA}=x \Omega \times \frac{2}{3} \mathrm{~mA} \\
\Rightarrow & x=3000 \times 1 \times \frac{3}{2}=4500 \Omega
\end{aligned}
$$
But the galvanometer resistance \(=50 \Omega\) Therefore the resistance to be added
$$
=(4500-50) \Omega=4450 \Omega \text {. }
$$
