= Solution
For an arbitrary <subgroup> $A$, the full inverse image of its image is $AK$, not generally $A$. In fact, $\theta(g)\in\theta(A)$ exactly when $g=ak$ for some $a\in A,k\in K$. Since $K$ is normal, $AK$ is a <subgroup>. By the <subgroup correspondence for a surjective group homomorphism>,
$$
[H:\theta(A)]=[G:AK].
$$
Multiplicativity of finite index gives
$$
[G:A]=[G:AK][AK:A].
$$
The map $k(K\cap A)\mapsto kA$ bijects the left <cosets> of $K\cap A$ in $K$ with those of $A$ in $AK$, since $AK=KA$. Consequently
$$
\boxed{[H:\theta(A)]=\frac{i}{[K:K\cap A]}.}
$$
This need not equal $i$. For example, take $\theta:\mathbb Z\to\mathbb Z/2\mathbb Z$ to be reduction modulo $2$ and $A=3\mathbb Z$. Then $[G:A]=3$, but $\theta(A)=H$ has index one. If $A$ contains $K$, however, the denominator is one and
$$
\boxed{K\leq A\ \Longrightarrow\ [H:\theta(A)]=i.}
$$
For finite $i$, equality holds exactly when $K\leq A$.
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