jmc

We must have $x>0$ for ${\log }_{1995}x$ to be defined. Thus, $x^2>0$, so both sides are positive. Then we may take logarithms of both sides, giving $${\log }_{1995}\left(\sqrt{1995}{x}^{{\log }_{1995}x}\right)={\log }_{1995}{x}^2$$or $${\log }_{1995}\sqrt{1995}+{\log }_{1995}{x}^{{\log }_{1995}x}={\log }_{1995}{x}^2.$$Using logarithm identities, this simplifies to $$\frac{1}{2}+{\left({\log }_{1995}x\right)}^2=2{\log }_{1995}x$$or $$({\log }_{1995}x{)}^2-2{\log }_{1995}x-\frac{1}{2}=0.$$By Vieta's formulas, the sum of the roots of this equation is $2$. That is, if $r_1$ and $r_2$ are the two values of $x$ that satisfy this equation, then $${\log }_{1995}{r}_1+{\log }_{1995}{r}_2=2.$$Now we have ${\log }_{1995}(r_1{r}_2)=2$, so $r_1{r}_2=1995^2=\boxed{3980025}$.