For maximum utilization the sender must send frames until ACK comes back. i.e., the frames must be sent until RTT.
$\text{Round trip propagation delay} = 550 ms$
$\text{Transmission time} = \frac{2000}{ 50} = 40 ms$
So, $RTT = 550 + 40 = 590 ms$.
Now, sender must send frames until $590 ms$. Time to send 1 frame = 40 ms. So, no. of frames required $=\left\lceil \frac{590}{40} \right \rceil = 15$. So, window size must be >= 15 frames.
The question doesn't mention maximum utilization and the fractional part in options doesn't make sense.