Voltage on coaxial line - calculation process
First, read this answer for a general description of what's going on.

In your specific example, at t = 0, 6V is applied to the transmission line, which initially has 0V everywhere along its length. Since the source has zero resistance, the full 6V appears across the 75&Omega; impedance of the line, and a current of 6V/75&Omega; = 80 mA flows into it. This wave propagates toward the terminator.

At t = 500 ns, this wave reaches the 150&Omega; termination. However, at 6V, only 40 mA would ordinarily flow through this termination, so there's an excess of 40 mA in the line. The voltage rises above 6V by enough to cancel this excess current in the line, keeping in mind that as the voltage rises, more current flows through the termination, too. For every volt above 6V that the voltage rises, 1/3 of the excess current flows through the 150&Omega; terminator, and 2/3 of it flows back into the 75&Omega; impedance of the line. Since we have an excess of 40 mA, the voltage rises until another 13.33 mA flows into the terminator and 26.67 mA flows back into the line. Since there's now a total of 53.33 mA flowing through the terminator, the voltage rises to 53.33 mA &times; 150 &Omega; = 8 volts. This wave (a step from 6V to 8V) propagates back toward the source. In general, you can say that this ratio of line impedance to termination impedance always reflects 1/3 of the incident wave back, with the same polarity.

At t = 1000 ns = 1.0 &micro;s, this wave reaches the source. However, since the source has zero impedance, the voltage there must remain at 6V. In effect, this creates a -2.0 V step that directly cancels the +2.0 V step created by the terminator. However, the current in the line was reduced by the terminator from 80 mA initially to 53.33 mA, and this is the value of the current throughout the line at this moment in time. But in order to create this new negative-going step, it is necessary for the current in the line to drop by another 2.0 V / 75 &Omega; = 26.67 mA. This step propagates toward the terminator, but now the current in the line behind it is reduced to a total of 26.67 mA. In other words, a zero-ohm termination (even at the source) reflects the incident wave at 100% amplitude, but the voltage is inverted and the incremental current is doubled.

At t = 1.5 &micro;s, this new step (&Delta;V = -2.0 V) reaches the terminator. But once again, the current and voltage do not match what the terminator requires (40 mA @ 6 V), and so the terminator once again reflects 1/3 of this wave back toward the source, making the line voltage 6.0V - 2.0V/3 = 5.333 V. This, by the way, is the answer to the original question, since this is the voltage that exists at the terminator from 1.5 &micro;s until the next reflection arrives from the source at 2.5 &micro;s.

These reflections keep going on, with the amplitude being reduced by 1/3 at each reflection at the terminator, until everything eventually settles down with the line voltage at 6V and 40 mA flowing through the line and the terminator.

The voltage at the termination goes like this:
$$
\begin{matrix}
time (\mu s) &amp; &amp; voltage \\
\hline
0.0 - 0.5 &amp;&amp; 0.000 V\\
0.5 - 1.5 &amp;&amp; 6.000 V + 1/3 \cdot 6.000 V = 8.000 V\\
1.5 - 2.5 &amp;&amp; 6.000 V - 1/9 \cdot 6.000 V =5.333 V\\
2.5 - 3.5 &amp;&amp; 6.000 V + 1/27 \cdot 6.000 V = 6.222 V\\
3.5 - 4.5 &amp;&amp; 6.000 V - 1/81 \cdot 6.000 V = 5.926 V\\
4.5 - 5.5 &amp;&amp; 6.000 V + 1/243 \cdot 6.000 V = 6.025 V\\
5.5 - 6.5 &amp;&amp; 6.000 V - 1/729 \cdot 6.000 V = 5.992 V\\
\end{matrix}
$$
... and so on.