Hint: To reduce the error, the value of \(S\) should of the order of \(R.\)

Step: Find the error in the resistance of the meter bridge experiment.
\(\Rightarrow R=S\left(\frac{l_1}{100-l_1}\right)=100\times \frac{2.9}{97.1}=2.98~ \Omega\)
In this problem, the balanced Wheatstone bridge is to be used.
Condition of balanced Wheatstone bridge: The bridge is said to be balanced if the ratio of the resistances in the same branch is equal \(\frac{R}{S}=\left(\frac{l_1}{100-l_1}\right)\)
Wheatstone bridge is an arrangement of four resistances that can be used to measure one unknown resistance of them in terms of rest. The percentage error in \( R\) can be minimised by adjusting the balance point near the middle of the bridge, i.e., when \(l,\) is close to \(50~\text{cm}.\) This requires a suitable choice of \(S.\)
Since, \(\frac{R}{S}=\left(\frac{l_1}{100-l_1}\right)\)
Since here, \(R: S=2.9: 97.1\) then the value of \(S\) is nearly \(33\) times that of \(R.\) In order to make this ratio \(1:1,\) it \(\text {is necessary to reduce the value of } S \text { almost } 1 / 33 \text { times, i.e., nearly } 3~ \Omega \text {. }\)
Hence, option (3) is the correct answer.