Let me be slightly more formal (Answer is 7 btw)
S = Ordered set of all triplets of vertices
$S=\{(i,j,k):i,j,k \epsilon V,i\neq j\neq k\} \\ |S| = \binom{8}{3}=56$
Now I'll make random variable X such that $X_p((i,j,k))=\begin{cases} 1 & \text{ if i,j,k form a cycle} \\ 0& \text{ otherwise } \end{cases}$
Expected value of X will be the sum of all expectations of $X_p$ (by linearity theorem), which will give us the expected 3-cycles (by definition of $X_p$ above
$E(X)=E(X_1)+E(X_2)...E(X_{56})\\=56(1.p+0.(1-p))=56p$
Here p is the probability of finding a cycle in 3 vertices $i\rightarrow j \rightarrow k$
To find p, note by independence of events that $p(i\rightarrow j \rightarrow k)=p(i\rightarrow j ).p(j\rightarrow k).p(k\rightarrow i )\\=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{8}$
Hence expected 3-cycles = 56p = 7 Q.E.D.