First off, let me say that you can find the answer to this question in Sage using the nauty generator. If you're going to be a serious graph theory student, Sage could be very helpful.
count = 0
for g in graphs.nauty_geng("20 180:180"):
count = count+1
print count
The answer is 4613. But, this isn't easy to see without a computer program.
At this point, perhaps it would be good to start by thinking in terms of of the number of connected graphs with at most 10 edges. Then, all the graphs you are looking for will be unions of these. You should be able to figure out these smaller cases. If any are too hard for you, these are more likely to be in some table somewhere, so you can look them up.
Connected graphs of order n and k edges is:
n = 1, k = 0: 1
n = 2, k = 1: 1
n = 3, k = 2: 1
n = 3, k = 3: 1
n = 4, k = 3: 2
n = 4, k = 4: 2
n = 4, k = 5: 1
n = 4, k = 6: 1
n = 5, k = 4: 3
n = 5, k = 5: 5
n = 5, k = 6: 5
n = 5, k = 7: 4
n = 5, k = 8: 2
n = 5, k = 9: 1
n = 5, k = 10: 1
.
.
.
n = 10, k = 9: 106
n = 10, k = 10: 657
n = 11, k = 10: 235
I used Sage for the last 3, I admit. But, I do know that the Atlas of Graphs contains all of these except for the last one, on P7.
As already noted in comments:
$\alpha$ and $\beta$ are explicitly specified to be "one-to-one correspondences", which is a way of saying "bijections". Since there is a bijection between $V_G$ and $V_H$, they have the same number of elements. Similarly, since there is a bijection between $E_G$ and $E_H$, they have the same number of elements.
Best Answer
If you consider labeled graphs, there are $n^2$ legitimate spots for an edge (if you allow loops). Since you accept the non-simple graphs (so parallel edges), each of the $k$ edges can be put in any of those spots. In the end, you get $(n^2)^k$ which is equal to $ n^{2k}$.