The speed of light in fiber is $3 \times 10^8 \: m/sec$
For $25$ kms distance the propagation delay is $10^{-4} \: sec$
$\begin{array}{ll} \text{Round Trip time} & = 2 \times \text{propagation delay} \\ & = 2 \times 10^{-4} \: sec \end{array}$
Given
$\text{Image size }= 4KB = 32768 \text{ bits}$
Transmission delay = Round trip time
$\begin{array}{lll} \Rightarrow & \frac{L}{B} & = 2 \times 10^{-4} \text{ sec} \\ \Rightarrow & B & = L/(2 \times 10^{-4}) \text{bits/sec} \\ \Rightarrow & B & = 32768 /(2 \times 10^{-4}) \text{ bits/sec} \\ \Rightarrow & B & = 16384 \times 10^4 \text{ bits/sec} \end{array}$