Ans: D
Since they are terms of a geometric sequence, then we have
$\begin{array}{rcl}
\dfrac{a_2}{a_1} & = & \dfrac{a_3}{a_2} \\
\dfrac{1}{8p^2} & = & \dfrac{27p}{1} \\
p^3 & = & \dfrac{1}{216} \\
p & = & \dfrac{1}{6}
\end{array}$
Therefore, the common ratio
$\begin{array}{cl}
= & \dfrac{a_2}{a_1} \\
= & \dfrac{1}{8(\frac{1}{6})^2} \\
= & \dfrac{9}{2}
\end{array}$
Hence, we have
$\begin{array}{rcl}
a_4 & = & a_1r^3 \\
a_4 & = & 8\left(\dfrac{1}{6}\right)^2 \times \left(\dfrac{9}{2}\right)^3 \\
a_4 & = & \dfrac{81}{4}
\end{array}$
