Logarithms are base 2. Use the inequality $|\sum_{a=1}^m \genfrac(){1pt}{2}{a}{p}|\leq \sqrt{p}\log p$, where $\genfrac(){1pt}{2}{a}{p} := a^{(p-1)/2}\mod p$ and p is a prime number, to prove that there exists some $1\leq a < p^{1/(2\sqrt{e})} (\log p)^2$ that is a quadratic nonresidue modulo p.
I know that the number of quadratic residues modulo p for an odd prime p is $(p+1)/2$ (if $x$ is a nonzero quadratic residue, then there are two distinct elements of $\mathbb{F}_p$ whose square is x so there is a two-to-one mapping from $\mathbb{F}_p^\times$ to the nonzero quadratic residues modulo p). Let $N$ be the number of quadratic non-residues among $1,2,\cdots, m.$ Let $X$ be the smallest quadratic non-residue. Then every quadratic non-residue has a prime divisor that's at least X, since the remainder of the non-residue must be at least X. So $N\leq \sum_{X\leq a<m} \genfrac(){1pt}{2}{a}{p}.$ Then one should be able to use the given inequality somehow.
Since no-one else has answered, I wil give you the strongest result I can. It can probably be strengthened by considering the speed of convergence in Mertens second theorem, but I don't have the time to work it all out.
Thm: Let $1<\alpha<\sqrt{e}$ and let $1-2\log\alpha>\varepsilon>0$. There exist an integer $N:=N_{\alpha,\varepsilon}$ such that for all $n\ge N$ and all primes $p$ the following is true: If $1,2,\ldots, n$ are all quadratic residues modulo $p$, we have $$ \left(1-2\log\alpha-\varepsilon\right)^{-1/\alpha} p^{1/2\alpha}(\log p)^{1/\alpha}\ge n $$
Let $1<\alpha<\sqrt{e}$ be any real number. For all $n\ge 1$, define $f(n)$ to be the number of integers $1\le k\le n^\alpha$ such that for all primes $q\mid k$, we have $q\le n$. Then $$ f(n)=\lfloor n^\alpha\rfloor-\sum_{n^\alpha\ge q\ge n}\left\lfloor\frac{n^{\alpha}}p\right\rfloor\ge -1+n^{\alpha}\left(1-\sum_{n^{\alpha}\ge q \ge n}\frac1q\right). $$ By Mertens' second theorem, $$ \lim_{n\to\infty}\sum_{n^{\alpha}\ge q \ge n}\frac1q = \lim_{n\to\infty}\left[\left(\sum_{n^\alpha\ge q}\frac1q-\log\log n^\alpha\right)+\log\alpha-\left(\sum_{n^\alpha\ge q}\frac1q-\log\log n\right)\right] = \log\alpha. $$ Therefore, $$ \lim_{n\to\infty}\frac{f(n)}{n^{\alpha}} = 1-\log\alpha. $$ Now let $n$ be a positive integer, $p$ be a prime, and assume $1,2,\ldots,n$ are all quadratic residues, then $$ \sqrt{p}\log p\ge \sum_{1\le a\le n^{\alpha}}\left(\frac ap\right) \ge 2f(n)-n^{\alpha} = n^{\alpha}\left(\frac{2f(n)}{n^{\alpha}}-1\right). $$ Let $\varepsilon > 0$. For $n$ large enough, we have $$ \begin{align*} \sqrt{p}\log (p)&\ge n^{\alpha}\left(1-2\log\alpha-\varepsilon\right)\\ \left(1-2\log\alpha-\varepsilon\right)^{-1/\alpha} p^{1/2\alpha}(\log p)^{1/\alpha}&\ge n. \end{align*} $$