In 25 consecutive natural numbers, the average of last 13 numbers is 39. What is the average of all 25 numbers?

Let the first of the $$25$$ consecutive natural numbers be $$n$$.
Hence the complete list is $$n,\;n+1,\;n+2,\;\dots,\;n+24$$.

The last $$13$$ numbers in this list are
$$n+12,\;n+13,\;n+14,\;\dots,\;n+24.$$

Because these $$13$$ numbers are consecutive, their average is the middle (7th) term of the block.
The 7th term is $$n+18$$.

Given that this average equals $$39$$:
$$n+18 = 39 \;\Rightarrow\; n = 21.$$

Therefore the $$25$$ numbers run from $$21$$ up to $$45$$.

For any sequence of consecutive numbers, the average of the entire set equals the average of the first and last terms:
$$\text{Average of all 25 numbers} = \frac{21 + 45}{2} = \frac{66}{2} = 33.$$

Hence the required average is $$33$$.

Option D which is: 33