The most important statement in question is
each task requires one unit of time
This shows that we can greedily choose the better task and that should give us the optimal solution. The best task would be the one with maximum profit. Thus we can sort the tasks based on deadline and then profit as follows:
$$\begin{array}{|l|l|l|l|l|l|l|l|l|}\hline \textbf{Task} & \text{$T_7$ } & \text{$T_2$}& \text{$T_9$} & \text{$T_4$} & \text{$T_5$} & \text{$T_3$} & \text{$T_6$} & \text{$T_8$} & \text{$T_1$} \\\hline \textbf{Deadline} & \text{2} & \text{2}& \text{3} & \text{3} & \text{4} & \text{5} & \text{5} & \text{7} & \text{7} \\\hline \end{array}$$
$$0 \overset{T_7}{-} 1 \overset{T_2}{-} 2 \overset{T_9}{-} 3 \overset{T_5}{-} 4 \overset{T_3}{-} 5 \overset{T_8}{-} 6 \overset{T_1}{-} 7$$
Thus, $T_4$ and $T_6$ are left.
So, maximum profit will not include those of $T_4$ and $T_6$ and will be $=15 + 20 +30 +18 + 16 + 23 + 25 = 147$
A is answer