Length of 1 time slot=Transmission time for 100 bits + propagation time (one-way).

Now

$T_t(100\,bits\,)=\frac{100bits}{10^7bps}=10\mu s$

$T_p=\frac{10^3}{2*10^8}=5 \mu s$

Cycle time or 1 slot time=$15 \mu s$

Now in a TDM based channel access method, each node gets dedicated bandwidth of $\frac{R}{N} bps$ where

R=Bandwidth of the broadcast medium

N=Total number of nodes in the network

For each node, the efficiency would be = $\frac{10}{15}=\frac{useful\,time}{total\,time}=\frac{2}{3}$

So, for each node the Throughput would be=Efficiency*Available Bandwidth=$\frac{2}{3}*\frac{10\,Mbps}{N}$

and this throghput is given in question i.e. $\frac{2}{3}$

$\frac{2}{3}*\frac{10\,Mbps}{N}=\frac{2}{3}\,Mbps$

**N=10. Answer**