Average ability conditioning on having accepted an offer
If $\alpha \sim U$, then how come there is no expectation in your profit function? The $\alpha$ is unknown and which $\alpha$-types the firm gets depends on salary $v$. This should be reflected in the profit function.
Next, your $n(v)$ seems to assume that $\alpha \sim U[0,1]$, but you set $\alpha \sim U[0,2]$. I assume this is a typo and I edited your question. Otherwise, you need to divide your $n(v)$ by two, because your density is $\frac{1}{2}$ instead of 1. This upperbound cancels out anyway.
For any uniformly distributed $\alpha&lt;\widehat v$, you have
$$\mathbb{E}[\alpha \vert \alpha \leq \widehat v] = \int_0^{\widehat v}  \alpha \frac{1}{\widehat v}d\alpha = \left[\frac{1}{2\widehat v}\alpha^2\right]_0^{\widehat v}= \frac{\widehat v}{2} $$
And in your case, it's ${\widehat v}=\sqrt{v/2} \Rightarrow \mathbb{E}[\alpha \vert \alpha \leq \sqrt{v/2}] = \sqrt{v/8}$. If $\alpha \sim U[0,x]$ then the density is $1/x$, but you also account for $\alpha&lt;\widehat v$ by dividing by $\widehat v/x$ such that $x$ cancels out.
Hence, your expected profit function is
$$\mathbb{E}[\pi(v)] = (\rho \sqrt{v/8}-v)\sqrt{v/2}.$$
Acceptance does not skew the distribution. All types below a cutoff accept, all others reject. Therefore, the distribution conditional on acceptance is uniform up to the cutoff.
If you replace your $n(v)$ by $n(v)/2$, taking account for the upperbound 2, you would get the same.