Classifying a set in the arithmetical hierarchy It is easy to see that not every recursive partial function has a recursive total extension:  consider a function which returns how long it takes $W_e$ to halt on input $e$.  If this function had a recursive total extension, we could use it to solve the halting problem.
What is the location of the set of all numbers $e$ such that $\phi_e$ has a primitive recursive extension (where $\phi_e$ is the $e$th  partial recursive function) in the arithmetical hierarchy?
 A: We can get an upper bound by just writing the sentence out, with quantifiers displayed clearly (and generally this upper bound is in fact exact, although that takes additional proof):
$\varphi_e$ has a primitive recursive extension iff there  is some $n$ such that
$$\mbox{For each $k$ and $s$, if $\varphi_e(k)[s]$ halts, then it equals $p_n$}$$ where $p_n$ denotes the $n$th primitive recursive function in some standard enumeration, and $\varphi_e(k)[s]$ represents the $e$th partial computable function run on input $k$ for $s$-many stages.
Now every primitive recursive function is total, so "it equals $p_n$" doesn't add any quantifier complexity (we can just wait to see the function halt); and "$\varphi_e(k)[s]$ halts", unlike "$\varphi_e(k)$ halts," is also $\Delta_1^0$ because of the time bound $s$. So the only quantifiers which apply here are the "$\exists n$" and the "$\forall k, s$" at the beginning; so this is at worst $\Sigma^0_2$.
And it's easy to show that in fact this index set is $\Sigma^0_2$-complete: given a $\Sigma^0_2$ sentence $\psi=\exists x\forall y\theta(x, y)$, we build a computable partial function $g=\varphi_{f(\psi)}$ which "gradually diagonalizes" against the primitive recursive functions, as follows:


*

*At the beginning of each stage, we'll have a current guess at a witness for $\psi$ being true. We begin with $m_0=0$.

*At stage $n$, we'll define $g(n)$.

*Specifically, at stage $n$, look for a counterexample $y\le n$ to "$\forall y\theta(m_n, y)$. If you find one, let $g(n)>p_i(n)$ for each $i\le n$ (where $p_i$ is the $i$th primitive recursive function in some fixed enumeration); otherwise, let $g(n)=p_{m_n}(n)$. If you find a counterexample, set $m_{n+1}=m_n+1$; otherwise, set $m_{n+1}=m_n$.
If $\psi$ holds, we eventually settle on the right witness $m$; at which point $g=p_m$ from then on. Since $g$ differs from a primitive recursive function at only finitely many points, $g$ is itself primitive recursive. Meanwhile, if $\psi$ fails, then for each $i$ there is some $n$ such that $g(n)>p_i(n)$ (just look for a "finds-a-counterexample" step after stage $i$); so $g$ differs from every primitive recursive function.
So $\psi$ holds iff $g=\varphi_{f(\psi)}$ can be extended to a primitive recursive function (in fact, is primitive recursive).

Now, you also ask about extendability to a total recursive function. This is different, since total recursive functions properly include primitive recursive functions; in particular, there is no effective enumeration of the total recursive functions (if there were we could diagonalize against it). 
So let's look at the new extendability property: $\varphi_e$ has a total recursive extension iff there is some $c$ such that $$\mbox{For all $k$ and $s$, there is some $t$ such that $\varphi_c(k)[t]$ halts and if $\varphi_e(k)[s]$ halts, it equals $\varphi_c(k)[t]$}.$$ We've gained a quantifier - specifically, the clause "there is some $t$ such that $\varphi_c(k)[t]$ halts". Now our upper bound is $\Sigma^0_3$. And, in fact, this is sharp - but that argument is a bit tedious so I'll leave it out.
