So far I can get to the fact that |G| $\geq$ 12 by Euler's Formula.
I'm having a hard time figuring out how to get to the $\frac{1}{5}|G|$ unmatched vertices.
I tried doing it by contradiction and started with picking a maximum matching and then assuming that there are more unmatched vertices but couldn't get too far with finding an issue.
I know that those unmatched vertices can't be adjacent to each other since then we could make a larger matching. So then they have to be adjacent to at least 5 different vertices in the matching since minimum degree is 5, but then I don't know where to go from here.
Edit: Some progress I think but needs some fleshing out
I got some more progress but still can't flesh out everything.
So I'm using Tutte's Theorem and picking a max matching $M$ where the matching will match a set $S$ into the odd components of $G$ ($o(G-S)$)
We need to consider 3 cases,
1.) Component has at least 5 vertices, this case is easy since it is clear that this will contribute exactly 1 unmatched vertex since by Tutte's Theorem we know that every component of $G-S$ is factor-critical, i.e.: There is a perfect matching of $n-1$ vertices in a component of size $n$.
2.) Component has size 1, then it has at least 5 connections to $S$.
3.) Component has size 3, now sure how to approach this case.
Then I want to show that the number of odd components of $(G-S)$ - $|S|$ is $\leq$ $\frac{1}{5}|G|$ because each component leaves exactly one vertex unsaturated. I'm guessing this is where I use Euler's Formula for Planar graphs?
Is it not like so? Only it seems like you could easily prove something stronger.
Let $k_i$ be the number of components of size $i$ in $G-S$ and $s=|S|$, so that $$n=s+k_1+3k_3+5k_5+...$$ Then there are $-s+k_1+k_3+k_5+... $ unmatched vertices and we want $$ -5s +5k_1 +5k_3 + 5k_5 +... \leq s+k_1+3k_3 + 5k_5 +...$$ So let's just show the weaker $ 4k_1+ 2k_3 \leq 6s$.
In fact $k_1 \leq s$ and $k_3 \leq s$, reason - consider the planar bipartite graph formed by the vertices of S and the vertices from the components of size 3 say, and the edges between these components and S, for the size 3 case each component offers at least 9 edges so that $9k_3 \leq 2(s + 3k_3) - 4$ and so $k_3 \leq s$ similarly $k_1 \leq s$.