IN CSMA/CD, We need to consider the longest path, Here $H_9$ and $H_{10}$ will have path length of 900 m and 1000 m from the hub, so total distance = 1900 m.
A signal needs to travel from $H_9$ to $H_{10}$ so total distance is 1900, along with that hub adds delay of 2.5 $\mu s$.While transmitting to $H_{10}$ if the collision happens near $H_{10}$, a collision signal will again take 2.5 $\mu s$ hub delay to come back to $H_{9}$. Hence a hub will add a total of 5 $\mu s$ delay overall.
Hence,
$T_t \geq 2*T_P + 2*T_{hub}$
where,
$T_{hub} = 2.5 \mu s$
Propagation time = $\frac{distance}{velocity} = \frac{1900}{2*10^8} = 9.5 * 10^{-6} = 9.5 \ \mu s$
$L \geq (2*T_P + 2*T_{hub})B = (2*9.5 \mu s + 2*2.5 \mu s)10* 10^{6}bps= 240 \ bits$ .
