Theory: Geometric Series is a series in which consecutive terms have the same ratio.

\(n^{th}\) term of the Geometric series is given by \(T{_n} = a*r^{n-1}\)
Where, a is the first term
r is the common ratio (or \(\frac{T{_2} }{ T{_1}}\) )
n is the term number

Given series is 1, 2, 4, 8, 16, 32
So, First term, a = 1
Common ratio, r = \(\frac{2}{1}\) = 2
And we need to find \(T{_{16}}, T{_{17}}, T{_{18}}\)
Using, \(T{_n} = a*r^{n-1}\). We get
\(T{_{16}} = 1*2^{16-1}\) = \(2^{15}\)
\(T{_{17}} = 1*2^{17-1}\) = \(2^{16}\)
\(T{_{18}} = 1*2^{18-1}\) = \(2^{17}\)

The sum of the 16th, 17th and 18th term = \(T{_{16}} + T{_{17}} + T{_{18}}\) = \(2^{15}\) + \(2^{16}\) + \(2^{17}\)
= \(2^{15}\) + \(2^{15+1}\) + \(2^{15+2}\)
= \(2^{15}\) + \(2^{15}\) *\(2^1\) + \(2^{15}\) *\(2^2\)
Taking \(2^{15}\) common we get
= \(2^{15}\) ( 1 + \(2^1\) + \(2^2\) ) = \(2^{15}\) * ( 1 + 2 + 4)
= 7*\(2^{15}\)

So, answer will be E
Hope it helps!

To learn more about the Geometric and Arithmetic series watch the below video