Packet Size = 1KB, Bandwidth = $10^9$ bps, Propagation Time = 15 ms
$\text{Transmission Time} = \frac{\text{Packet Size}}{\text{Bandwidth}} \\= \frac{1 \text{ KB}}{ 10^9 \text{ bps}} \\ = \frac{1024 \times 8 }{10^9 \text{ bps}} \approx 0.008 \text{ ms}.$
Sender Utilization = Fraction of Time Sender is busy (Same as link utilization which is the fraction of time the link is carrying useful data)
In Stop-and-wait, sender can retransmit only when ACK arrives for the send data which requires 1 transmission time for the data packet, 1 propagation delay for the data packet to reach the receiver, 1 propagation delay for the ACK to reach the sender. (ACK normally being small its transmission time can be ignored unless given in question). So,
$\text{Utilization} = \frac{\text{Transmission Time}}{\text{ Transmission Time } + 2 \times \text{Propagation Time}} \\= \frac{0.008}{0.008 + 2 \times 15} \\= \frac{0.001 }{30.008} \\= 0.000266 = 0.026 \%.$