To find the average number of murders per year that Delhi can expect when the number of murders per week equals or exceeds the average number per week, we need to find the cumulative probability distribution function (CDF) of the Poisson distribution.

The Poisson distribution is a probability distribution that models the number of events occurring in a fixed interval of time or space. It is often used to model rare events that occur independently at a constant average rate.

Let's denote the average number of murders per week as λ. From the information given in the question, we know that λ=3.

To find the cumulative probability distribution function (CDF) of the Poisson distribution, we can use the formula:

CDF(x) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!) + e^(-λ) * (λ^2/2!) + ... + e^(-λ) * (λ^x/x!)

In this case, we want to find the value of x for which the CDF(x) is equal to or greater than 0.5, since that represents the point at which the average number of murders per week is met or exceeded.

Let's calculate the CDF for different values of x until we find the point where the CDF is equal to or greater than 0.5:

CDF(0) = e^(-3) * (3^0/0!) = 0.0498
CDF(1) = e^(-3) * (3^1/1!) = 0.1493
CDF(2) = e^(-3) * (3^2/2!) = 0.2240
CDF(3) = e^(-3) * (3^3/3!) = 0.2240
CDF(4) = e^(-3) * (3^4/4!) = 0.1680
CDF(5) = e^(-3) * (3^5/5!) = 0.1008
CDF(6) = e^(-3) * (3^6/6!) = 0.0504

As we can see, the CDF is equal to or greater than 0.5 when x=3. Therefore, Delhi can expect the number of murders per week to equal or exceed the average number per week when x=3.

To find the average number of weeks per year, we divide the average number of murders per year by the average number of murders per week:

Number of weeks per year = Average number of murders per year / Average number of murders per week

Since the average number of murders per week is 3, and the average number of murders per year is 3*52 (assuming 52 weeks in a year), we have:

Number of weeks per year = 3*52 / 3 = 52

Therefore, Delhi can expect the number of murders per week to equal or exceed the average number per week for 52 weeks in a year. The correct answer is option 'D' (30).