$T_{p}$=270ms=0.27s (Standard for geostationary satellites)
$\therefore$RTT=0.27*2=0.54s
Bandwidth(B)=50Mbps=$50\times10^{6}bps (\because 1Mbps=10^{6}bps)$
Bandwidth*Delay product=B*RTT=$50*10^{6}*0.54\ bits=27000000\ bits$ (This answers the first part)
Bandwidth-Delay product means maximum number of bits that can be present in the network at any given point in time.
Packet size(L) given=1500 B = $1500\times8\ bits(\because 1Byte =8bits)$
Now,if we manage to find maximum number of packets that can be present in the network,we also find the optimal window size.
N=$B*RTT\over L$=$27000000\over{1500*8} $=2250(Optimal window size)