Let the principal be Rs P. Then,
101.50 = P\(\Big[\big(1+\frac{3}{100}\big)^2-1\Big]\) = P\(\Big[\big(\frac{103}{100}\big)^2-1\Big]\)
\(\Rightarrow\) 101.50 = P\(\Big[\frac{(103)^2-(100)^2}{10000}\Big]\) = P\(\Big[\frac{(103+100)(103-100)}{10000}\Big]\)
\(\Rightarrow\) P= \(\frac{101.50\times10000}{203\times3}\) = Rs\(\frac{5000}{3}\)
\(\therefore\) S.I = \(\frac{5000\times3\times2}{3\times100}\) = Rs 100.