Make the obvious independence assumptions, and proceed along the lines described by Michael Chernick. Since everything has normal distribution, there is no need to make a normal approximation.
Let $X_i$ be the weight of the $i$-th car, and let $S=\sum X_i$. Then $S$ has normal distribution with mean $3n$ and variance $(0.3)^2 n$. Let $Y=S-W$. Then $Y$ has normal distribution with mean $3n-400$, and variance $v_n=(0.3)^2 n+(40)^2$.
We want to find smallest $n$ such that $\Pr(Y \gt 0) \gt 0.1$. To do that, we solve the equation $\Pr(Y \gt 0) = 0.1$. We have
$$\Pr(Y\gt 0)=\Pr\left(\frac{Y-(3n-400)}{\sqrt{v_n}}\gt \frac{400-3n}{\sqrt{v_n}}\right)=\Pr\left(Z \gt \frac{400-3n}{\sqrt{v_n}}\right),$$
where $Z$ is standard normal.
But the point $c$ such that $\Pr(Z \gt c)=0.1$ is approximately given by $c=1.28$.
So we need to solve the equation
$$\frac{400-3n}{\sqrt{(0.3)^2 n+(40)^2}}=1.28.\tag{$1$}$$
Multiply through by the denominator, and square both sides. We get a quadratic equation in $n$, with somewhat messy coefficients. Solve as usual. We want the smaller root. (The larger root is extraneous, introduced when we squared both sides.)
Or else solve Equation $(1)$ directly by using the "Solve" button on a calculator, or any one of the many programs that do such things.
Our root is roughly $116.2$. So the number of cars such that the probability of damage is $\gt 0.1$ is $117$.
Remark: If we think about it, we can see that there is no need to solve a quadratic. For it is clear that the number of cars must be less than $400/3$. So the contribution of the cars to the variance of $Y$ is at most $(400/3)(0.09)$, which is $12$. The contribution of the bridge strength variance to the variance of $Y$ is $1600$, whose bulk makes $12$ negligible. Thus we can assume that $Y$ has standard deviation $40$, and for all practical purposes we want to solve the linear equation
$$400-3n=(1.28)(40).$$
The number of passengers that show up has a binomial distribution with n=160, p=0.9.
To calculate the expected refund, you would add the expected refund from overbooking to the expected refund from no-shows.
The expected refund from overbooking would be: $$\sum_{k=151}^{160} P(X=k)*500(k-150) $$
The expected refund from no-shows would be $75*0.1*160=1200$
To calculate the standard deviation you would need to also calculate $E[Y^2]$ which is the sum of $E[Y^2]$ for overbooking and $E[Y^2]$ for no-shows (Y=amount refunded).
For the overbooking, $E[Y^2]$ would be: $$\sum_{k=151}^{160} P(X=k)*[500(k-150)]^2 $$ and for the no-shows, it would be $75^2*[(160*.1)(1-.1+160*.1)$
The standard deviation is $\sqrt{E[Y^2]-E[Y]^2}$
Best Answer
You are correct that the answer is very nearly 0, but (even after @Drew's nice edit) it does not seem you are at all clear on how to get the answer.
Here is an outline:
1) The total weight $T = X_1 + X_2 + \cdots + X_{25}$ is normally distributed. Figure out its mean $\mu_T$ and its SD $\sigma_T.$
2) Standardize: $P(T > 6000) = P\left(Z = \frac{T-\mu_T}{\sigma_T} > \frac{6000 - \mu_T}{\sigma_T}\right).$
3) Evaluate the quantity $\frac{6000 - \mu_T}{\sigma_T}$, and then the probability, using standard normal tables or software.
Below is a graph of the density function of the distribution of $T$, Your answer is the area beneath the curve to the right of the vertical red line.