To build on top of Andy aka's answer.
$$|V_{out}|=|V_{in}|×\frac{\omega RC}{\sqrt{1+\omega²R^2C^2}}$$
$$H(\omega)=\frac{\omega RC}{\sqrt{1+\omega^2R^2C^2}}$$
\$H(\omega)\$ will give us the absolute value of the filter, since it's already absolute-valued by Andy aka. I'm aware that a more normal way to write it would be \$|H(S)|\$, but that's our \$H(\omega)\$. So we're cool.
There's a couple of free parameters, but I'd start with \$R\$ and set it to \$1kΩ\$ because that's a reasonable value, not too high and not too small. \$\omega\$ is \$2\pi×50×10^3\$. The last parameters we get to choose is the resulting amplitude, how much do you really want?
Do you want \$\frac{\sqrt{2}}{2}≈70\%\$? Or perhaps you want \$95\%\$? I'll assume you want \$95\%\$.
So what we want is the last parameter, \$C\$
After some fiddling, you'll get this equation:
$$C=\frac{\frac{95}{\sqrt{100^2-95^2}}}{\omega R}$$
If we plug in our numbers it will look ugly like this:
$$C=\frac{\frac{95}{\sqrt{100^2-95^2}}}{(2\pi×50×10^3) (10^3)}=9.683nF$$
The cutoff frequency (-3dB) will be at:
$$\omega_c=\frac{1}{RC}=\frac{1}{(10^3)(9.683×10^{-9})}=103krad/s=16.4kHz$$
You tell me if that's good enough for you. But that's what you get with a first order RC HP filter.
Here's a schematic if you want to mess around or if you don't believe me.

And here's the link.