Data count register gives the number of words the DMA can transfer in a single cycle..
Here it is $16$ bits.. so max $2^{16}$ words can be transferred in one cycle..
Since memory is byte addressable.. $1 \text{ word}=1\;\text{byte}$
so $2^{16}$ bytes in $1$ cycle..
Now for the given file..
File size $=29154\ \textsf{KB} = 29154\times 2^{10}\ \textsf{B}$
$1$ cylce $\rightarrow$ DMA transfers $2^{16}\ \textsf{B}$
i.e
$1\ B$ transfered by DMA $\rightarrow \dfrac{1}{2^{16}}$ cycles.
Now, for full file of size $29154\ \textsf{KB},$
Minimum number of cylces $=\dfrac{(29154\times 2^{10}\ B)}{2^{16}}= 455.53$
But number of cylces is asked so $455.53\rightarrow 456.$