Ans: A
$\begin{array}{rcl}
x & = & kt^a \\
\log_3 x & = & \log_3 (kt^a) \\
\log_3 x & = & a\log_3 t +\log_3 k
\end{array}$
$\begin{array}{rcl}
x & = & kt^a \\
\log_3 x & = & \log_3 (kt^a) \\
\log_3 x & = & a\log_3 t +\log_3 k
\end{array}$
Note that the $y$ intercept of the graph is $-4$. Then we have
$\begin{array}{rcl}
\log_3 k & = & -4 \\
k & = & 3^{-4} \\
& = & \dfrac{1}{81}
\end{array}$
