Let $X$ be a nonsingular variety with $Y$ of codimension $r\geq 2$ nonsingular subvariety. Let $\pi:\widetilde{X}\rightarrow X$ be the blow of $X$ along $Y$ and $Y'=\pi^{-1}(Y)$ the exceptional divisor.
In the hint for the exercise, Hartshorne suggests that we prove that $\omega_Z\cong \mathcal{O}_Z(-q-1)$ where $Z$ is the fibre of $Y'$ over a closed point $y\in Y$. This is needed to prove the main result of the problem.
I am having trouble actually finishing off this sheaf computation. From Hartshorne Exercise II.8.3(b) and the fact that $Z\cong \operatorname{Spec}k(y)\times_k Y'$, we know that $$ \omega_Z\cong p_1^*\omega_y\otimes p_2^*\omega_{Y'}. $$ From the first of the hint, we know $\omega_{Y'}\cong \pi^*\omega_X\otimes \mathcal{O}_{Y'}(-q-1)$. Also, it is not hard to see that $\omega_y=\mathcal{O}_y$ so that means $p_1^*\omega_y\cong \mathcal{O}_Z$. This reduces the computation to $$ \omega_Z\cong p_2^*(\pi^*\omega_X\otimes\mathcal{O}_{Y'}(-q-1)). $$ Since $(-)^*$ distributes over tensor products, we get $$ \omega_Z\cong p_2^*\pi^*\omega_X\otimes p^*\mathcal{O}_{Y'}(-q-1)\cong p_2^*\pi^*\omega_X\otimes \mathcal{O}_Z(-q-1). $$ To get the claimed isomorphism, I should be able to prove $p_2^*\pi^*\omega_X\cong \mathcal{O}_Z$. However I am having trouble actually doing this.
Any suggestions would be appreciated! I am actually not too good with computations involving the canonical sheaf so suggestions on how to think about the canoniacl sheaf is also appreciated.
Edit: Here is the full text of the problem.
You have one big error here at the start which makes solving the rest of the problem difficult. $Z$ is not $\operatorname{Spec} k(y)\times_k Y'$ unless $y$ is a point - the former is a copy of $\Bbb P^{r-1}$ while the latter is of dimension $\dim X - 1$. Let's get the correct diagram out there first: $\def\w{\omega} \def\PP{\mathbb{P}} \def\ZZ{\mathbb{Z}} \def\cI{\mathcal{I}} \def\cL{\mathcal{L}} \def\cM{\mathcal{M}} \def\cN{\mathcal{N}} \def\cO{\mathcal{O}} \def\Pic{\operatorname{Pic}}$
$$\require{AMScd} \begin{CD} Z @>{p_2}>> \widetilde{X}\\ @V{p_1}VV @VV{\pi}V \\ \{y\} @>{i}>> X \end{CD}$$
Following the hint, we can write $\w_{\widetilde{X}}\cong \pi^*\cM\otimes\cL(qY')$ for some $\cM$ a line bundle on $X$ and $q\in\ZZ$. Since $(\pi^*\cM)|_{\widetilde{X}\setminus Y'}\cong \cM|_{X\setminus Y}$ and $\Pic X\setminus Y\cong \Pic X$ (corollary II.6.16 plus the corresponding statement on class groups from before), we see that $\cM$ is determined by its behavior on $\widetilde{X}\setminus Y' \cong X\setminus Y$, and therefore $\cM\cong\w_X$.
Continuing with the hint, we have that $\w_{Y'}\cong \w_{\widetilde{X}} \otimes \cL(Y') \otimes \cO_{Y'}$ by proposition II.8.20. By our assumption that $\w_{\widetilde{X}}\cong \pi^*\w_X\otimes \cL(qY')$, we see that $\w_{Y'}\cong\pi^*\w_X\otimes\cL((q+1)Y')\otimes\cO_{Y'}$. As $\cL(Y')\cong \cI_{Y'}^\vee$ by proposition II.6.18 and $\cI_{Y'}\cong \cO_{\widetilde{X}}(1)$ by the proof of proposition II.7.13, we obtain $\w_{Y'}\cong\pi^*\w_X\otimes\cO_{\widetilde{X}}(-q-1)\otimes\cO_{Y'}$.
Taking a closed point $y\in Y$ and letting $Z=\{y\}\times_Y Y'\cong\PP^{r-1}$, we see that $\w_Z\cong \w_{Y'}\otimes \bigwedge^{\dim Y} \cN_{Z/Y'}$ by proposition II.8.20. This gives us that $$\w_Z\cong \pi^*\w_X\otimes \cO_{\widetilde{X}}(-q-1) \otimes \cO_{Y'} \otimes \bigwedge^{\dim Y} \cN_{Z/Y'}.$$ Since $\cN_{Z/Y'}\cong\pi^*\cN_{\{y\}/Y}$ because they're both free sheaves of rank $\dim Y$ (use theorem II.8.24 to see that locally $Y'\to Y$ is $\PP^{r-1}\times Y\to Y$ then apply exercise II.8.3 for $\cN_{Z/Y'}$; note that $\{y\}$ is a point and the only vector bundles on a point are free) and exterior powers commute with pullback, our expression simplifies to $$\w_Z\cong \pi^*(\w_X\otimes \bigwedge^{\dim Y}\cN_{\{y\}/Y}) \otimes \cO_{\widetilde{X}}(-q-1),$$ which by another application of proposition II.8.20 is just $$\w_Z\cong \pi^*\w_{\{y\}} \otimes \cO_{\widetilde{X}}(-q-1).$$ As $\w_{\{y\}}\cong \cO_{\{k\}}$ and $\pi^*\w_{\{k\}}\cong\cO_Z$, we see that $\w_Z\cong \cO_Z(-q-1)$. On the other hand, $Z\cong \PP^{r-1}$ so $\w_Z\cong\cO_Z(-r)$ and therefore $q=r-1$.