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Explanation:

The last number is 3¹⁰⁰.

Sum of the remaining numbers = S = 1 + 3 + 3² + … + 3⁹⁹   …(1)

This is a GP with a = 1; r = 3 and n = 100

∴ S = $\frac{1\times (3^{100}-1)}{3-1}$
 
∴ Required ratio = $\hspace{0.17em}\frac{3^{100}}{\frac{1\times (3^{100}-1)}{2}}$ = $\hspace{0.17em}\frac{2}{1-{\displaystyle \frac{1}{3^{100}}}}$
 
∵ 1/3¹⁰⁰ is approximately zero, hence it can be ignored.
 
∴ Ratio = 2
 
Hence, option (b).