Correct answer: A. \(\frac{5}{3}\)
Explanation: The direct answer is option A: \(a_6=\frac{5}{3}\). In a geometric progression, moving from one term to a later term means multiplying by the common ratio once for each step. From the third term to the sixth term there are three steps, so \(a_6=a_3r^3\). Substitute the given values: \(a_6=45\left(\frac13\right)^3=45\times\frac1{27}=\frac{45}{27}=\frac53\). Therefore option A is correct. Option B, 5, would result from an incomplete or incorrect division and is too large for three multiplications by \(\frac13\). Option C, 15, is what one might get after dividing only once, so it ignores two of the three steps. Option D, 135, increases the value and would be inconsistent with a ratio smaller than 1. The terms move downward: \(a_4=15\), \(a_5=5\), and \(a_6=\frac53\), which confirms the result. Memory cue: between term numbers \(m\) and \(n\), the power is \(n-m\), not \(n\) itself.