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.