Let $$\Phi(x):=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-\frac{t^2}2}\:{\rm d}t\;\;\;\text{for }x\in\mathbb R$$ and $$f(x):=2x^2\Phi\left(-\frac{cx}2\right)\;\;\;\text{for }x\in\mathbb R$$ for some $c>0$.
How can we compute the maximum of $f$ on $(0,\infty)$ (at least approximately)?
A theoretical derivation of the maximum seems to be complicated. So, what can we do?
$$\Phi(x):=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-\frac{t^2}2}\,dt=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ $$f(x)=2x^2\Phi\left(-\frac{cx}2\right)=x^2 \left(1-\text{erf}\left(\frac{c x}{2 \sqrt{2}}\right)\right)$$
Let $\frac{c x}{2 \sqrt{2}}=y$ to make $$f(y)=\frac{8 }{c^2}y^2 (1-\text{erf}(y))$$ So, we need to find the extremum of $$g(y)=y^2 (1-\text{erf}(y))\implies g'(y)=2 y \left(\text{erfc}(y)-\frac{e^{-y^2} y}{\sqrt{\pi }}\right)=0$$ that is to say the root of the transcendental equation $$\text{erfc}(y)-\frac{e^{-y^2} y}{\sqrt{\pi }}=0$$
Graphing, we can see that the solution is close to $y=1$. Using Taylor expansion, we have $$\text{erfc}(y)-\frac{e^{-y^2} y}{\sqrt{\pi }}=\left(\text{erfc}(1)-\frac{1}{e \sqrt{\pi }}\right)-\frac{y-1}{e \sqrt{\pi }}+\frac{3 (y-1)^2}{e \sqrt{\pi }}+O\left((y-1)^3\right)$$ Ignoring the higher order terms, solving the quadratic in $(y-1)$ gives, as an estimate, $$y=\frac{1}{6} \left(7-\sqrt{13-12 e \sqrt{\pi } \text{erfc}(1)}\right)\approx 0.837293$$ making, as an estimate, $$x=\frac{\sqrt{2} \left(7-\sqrt{13-12 e \sqrt{\pi } \text{erfc}(1)}\right)}{3 c}\approx \frac{2.36822}{c}$$
If we want to polish the root, let us use Newton method for the zero of function $$h(y)=\text{erfc}(y)-\frac{e^{-y^2} y}{\sqrt{\pi }}$$ $$h'(y)=\frac{e^{-y^2} \left(2 y^2-3\right)}{\sqrt{\pi }}$$ Starting with our fist estimate, the iterates will then be $$\left( \begin{array}{cc} n & y_n \\ 0 & 0.8372929290 \\ 1 & 0.8418425951 \\ 2 & 0.8418822134 \\ 3 & 0.8418822164 \end{array} \right)$$ which leads to the solution $$x=\frac{2.3812024967}{c}$$ as already given by metamorphy in comments.