Since I was the one who answered the previous question you are using as a model for this question, I would like to start by answering this part of your question:
How the level payment was found for the second table?
The main insight is to plan for the increasing annuities to have the same term (number of payments). Since the first payment is $1000$ and each subsequent payment increases by $50$, we can't just write $\require{enclose} 1000(Ia)_{\enclose{actuarial}{n}i}$ because this would correspond to a cash flow of $1000, 2000, 3000$, etc. And we can't write $1000 a_{\enclose{actuarial}{n}i} + 50(Ia)_{\enclose{actuarial}{n}i}$ because then the first payment is $1050$, not $1000$. So we have to write the first payment as the sum of a level annuity of $950$ plus an increasing annuity where the increment is $50$ per payment. That's where $950$ comes from. Similarly, because the second series of payments begins at $1550$ and increases by $100$ each period, we have to write this as a level annuity of $1450$ plus an increasing annuity with increment $100$, in order to get the total to match.
That said, your list of payments is correct, but the way you structured the table is not. For each row, the last column must be the sum of the other columns. When you chose $2000$ for the level portion, that already exceeds the total amount scheduled to be paid, which would mean the decreasing column would need to be negative, not positive; for instance:
$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\
\hline
1 & 1000 & 0 & 0 & 1000 \\
2 & 1000 & 100 & 0 & 1100 \\
3 & 1000 & 200 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
11 & 1000 & 1000 & 0 & 2000 \\
12 & 2000 & 0 & \color{red}{-}200 & 1800 \\
13 & 2000 & 0 & \color{red}{-}400 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 2000 & 0 & \color{red}{-}1\color{red}{8}00 & 200 \\
\end{array}$$
This would then correspond to the equation of value (corrections in red)
$$\require{enclose}
\begin{align}
PV
&= 1000(v + v^2 + \cdots + v^{11}) + 2000(v^{12} + v^{13} + \cdots + v^{20}) + (100v^2 + 200v^3 + \cdots + 1000v^{11}) \\ &\qquad \color{red}{-} (200v^{12} + 400v^{13} + \cdots + 1\color{red}{8}00v^{20}) \\
&= 1000 a_{\enclose{actuarial}{11} i} + 2000 v^{11} a_{\enclose{actuarial}{9} i} + 100 v (Ia)_{\enclose{actuarial}{\color{red}{10}}i} \color{red}{-} 200 v^{11}(\color{red}{I}a)_{\enclose{actuarial}{9}i}.
\end{align}$$
Why is the last annuity increasing and not decreasing? Because that's how you chose to write it in the table! A decreasing annuity is the same as a level annuity minus an increasing annuity. Notice how in the table, the payments in the column "Increasing 2" are negative but are increasing in magnitude.
Instead of doing it this way, I would suggest writing the table like this:
$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing} & \text{Decreasing} & \text{Total} \\
\hline
1 & 900 & 100 & 0 & 1000 \\
2 & 900 & 200 & 0 & 1100 \\
3 & 900 & 300 & 0 & 1200 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
10 & 900 & 1000 & 0 & 1900 \\
11 & 0 & 0 & 2000 & 2000 \\
12 & 0 & 0 & 1800 & 1800 \\
13 & 0 & 0 & 1600 & 1600 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 0 & 0 & 200 & 200 \\
\end{array}$$
By now, you should see how this works. Since the first payment is $1000$ but increases only by $100$ each period, we split this into a level annuity of $1000 - 100 = 900$, plus an increasing annuity with increment $100$. This structure holds for $10$ payments, until the $10^{\rm th}$ payment, which is $1900$. Then on the $11^{\rm th}$ year, we begin paying a $10$-year decreasing annuity where the decrement is by $200$ each period, which automatically takes care of the requirement that the $11^{\rm th}$ payment is $2000$, since $10 \times 200 = 2000$. Using this table, you should easily be able to construct the equation of value. In both cases, you should get an answer of about $13117.4$.
Finally, to help with your understanding, suppose the last ten payments are not decreasing by $200$, but by $50$. So the list of payments becomes
$$1000, 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, \\
2000, 1950, 1800, 1750, 1700, 1650, 1600, 1550, 1500, 1450.$$
If you were to modify my version of the table, how would you do it? Obviously, you don't need to change the first ten rows, only the last ten rows. What might you write in the "Level payment" column? What would you write in the "Decreasing" column? What is the resulting equation of value?
Best Answer
The point is that, because of the claim limit, you do not observe the values above $200$, just the numbers of these censored values. What those high values would have been if they had not been censored could affect the likelihood, but you do not know them
Given $\theta$, the probability density for an observation of $x_i$ below $200$ is $\frac1\theta e^{-x_i/\theta}$, while the probability (not density) for an observation $x_j$ above $200$ is $e^{-200/\theta}$
That makes the likelihood for $\theta$ from the observations below $200$ proportional to $\frac1\theta^{1114} e^{-142752/\theta}$ and the likelihood from the observations above $200$ proportional to $e^{-200N/\theta}$. Multiply these together and you get an overall likelihood proportional to $\frac1\theta^{1114} e^{-(142752+200N)/\theta}$. That likelihood is maximised when $\theta = \frac{142752+200N}{1114}$
We are told that $\hat{\theta}= 168$, which suggests $N = \frac{1114 \times 168 - 142752}{200} = 222$