We have $a_{n+1}^{2}-a_{n}^{2}=1$ for all natural number $n$. Using a telescopic sum we obtain
$$n_{0}=\sum_{k=n_{0}}^{2 n_{0}-1}\left(a_{k+1}^{2}-a_{k}^{2}\right)=a_{2 n_{0}}^{2}-a_{n_{0}}^{2}=9 a_{n_{0}}^{2}-a_{n_{0}}^{2}=8 a_{n_{0}}^{2} .$$
Hence
$$a_{n_{0}}^{2}=\frac{n_{0}}{8}$$
On the other hand,
$$n_{0}-1=\sum_{k=1}^{n_{0}-1}\left(a_{k+1}^{2}-a_{k}^{2}\right)=a_{n_{0}}^{2}-a_{1}^{2}=\frac{n_{0}}{8}-a_{1}^{2},$$
or equivalently
$$a_{1}^{2}=\frac{8-7 n_{0}}{8}$$
Since $a_{1}^{2} \geq 0$, we deduce that $n_{0}=1$ and $a_{1}^{2}=\frac{1}{8}$.
Therefore
$$45=\sum_{k=1}^{45}\left(a_{k+1}^{2}-a_{k}^{2}\right)=a_{46}^{2}-a_{1}^{2}=a_{46}^{2}-\frac{1}{8},$$
which leads to
$$a_{46}=\frac{19 \sqrt{2}}{4} .$$