Here's some ideas. Good way to start with something like this is to first rule out what the groups $G$ and subgroups $K$ and $N$ cannot be like.
First of all, you want to find a group $G$ that actually has non-normal subgroups. So we can rule out $G$ abelian, there are definitely no examples there since every subgroup of an abelian group is normal.
Also, you want $K \neq N$ because if $K = N$ and $N$ is normal, then so is $K$. So $K$ has to be a proper subgroup of $N$ and of course $K \neq \{1\}$, so $N$ must have order $\geq 4$. Since $N$ also be a proper subgroup (else $N = G$ and $K$ is normal in $G$), the group $G$ must have order $\geq 8$.
There also one more thing you can notice. The group $N$ cannot be cyclic, because if $N$ is a normal cyclic subgroup of $G$, then every subgroup of $N$ is also normal in $G$.
So we start looking at non-abelian groups. The smallest one is $S_3$, but it's too small to have an example in it. The next ones are of order $8$, quaternion group $Q_8$ and dihedral group $D_8$. Every subgroup of $Q_8$ happens to be normal, but it turns out there is an example in $D_8$.