Ans: C
Since $ABCD$ is a rectangle, then we have
$\begin{array}{rcl}
AB & = & \dfrac{170}{2} -BC \\
AB & = & 85 -34 \\
AB & = & 51 \text{ cm}
\end{array}$
In $\Delta ABE$,
$\begin{array}{rcll}
BE^2 & = & AB^2 -AE^2 & \text{(Pyth. Thm.)} \\
BE^2 & = & 51^2 -24^2 \\
BE & = & 45\text{ cm}
\end{array}$
Note that $\angle ABE = \angle BCF$. Then in $\Delta BCF$, we have
$\begin{array}{rcl}
\sin \angle BCF & = & \dfrac{BF}{BC} \\
BF & = & BC \sin \angle ABE \\
BF & = & BC \times \dfrac{AE}{AB} \\
BF & = & 34 \times \dfrac{24}{51} \\
BF & = & 16\text{ cm}
\end{array}$
Hence, we have
$\begin{array}{rcl}
EF & = & EB +BF \\
EF & = & 45 +16 \\
EF & = & 61 \text{ cm}
\end{array}$
