Answer is 2500 bytes per second.
Throughput is number of bytes we are able to send per second.
Calculate the transmission time of sender $T_{t(\text{Send})},$ calculate one way propagation delay $T_p,$ Calculate the transmission time of receiver $T_{t\text{(Recv)}}$
We get $T_{t(\text{Send})}$ here as $\dfrac{1}{10}$ seconds,
$T_p$ as $\dfrac{1}{10}$ seconds( given in question as $100\ \text{ms}$ ),
$T_{t\text{(Recv)}}$ as $\dfrac{1}{10}$ seconds.
So , total time taken to send a frame from sender to destination,
$=T_{t(\text{Send})}+2\times T_p+T_{t\text{(Recv)}}=\dfrac{4}{10}$ seconds
So, we can send $\text{1000 bytes}$ (frame size) in $\dfrac{4}{10}$ seconds.
in $\text{1 second}$, we can send $\text{2500 bytes}$. So throughput is $\text{2500 bytes}$ per second.