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.
Hint In any planar graph with 6 vertices you have
$$e \leq 3v-6=12 .$$
This proves that any graph with at least 13 edges is non-planar.
All you need to do now, is check the graphs with 12 or less edges.
Since $K_5$ has $10$ edges and $K_{3,3}$ has $9$ edges, any non-planar graph has at least 9 edges.
9 edges: Your graph must be $K_{3,3,}$ (Why?)
10 edges: Your graph must be $K_{3,3,}$ with an extra edge or $K_5$ with an isolated vertex. (Why?)
11 edges and 12 edges: If your graph contains $K_{3,3}$, it is easy: it must be $K_{3,3}$ with enough extra edges.
For $K_5$ you need to look at two different situations: your graph contains $K_5$ as a subgraph, or the sixth vertex is the vertex of degree $2$ in a subdivision of $K_5$.
Best Answer
Let $G$ be a $4$-regular graph on $7$ vertices, and let $\overline{G}$ be the complement of $G$. $\overline{G}$ is regular; what is its degree (what you called order in your question)? What do you know about regular graphs of that degree? They’re very easy to count, and since $G_1$ is isomorphic to $G_2$ iff $\overline{G_1}$ is isomorphic to $\overline{G_2}$, counting the complements is as good as counting the graphs themselves.
(Note that the answer depends greatly on whether you’re counting labelled or unlabelled graphs. Also, I’m assuming that you’re looking only at simple graphs, i.e., without loops or multiple edges.)
Added:
To see that counting the complements is good enough, let $\mathscr{G}_n$ be the set of all simple graphs on $n$ vertices, and let $\varphi:\mathscr{G}_n\to\mathscr{G}_n:G\mapsto\overline{G}$ be the map that takes each graph in $\mathscr{G}_n$ to its complement. Then show that $\varphi$ is a bijection, and that $G\in\mathscr{G}_n$ is $k$-regular iff $\varphi(G)=\overline{G}$ is $(n-1-k)$-regular.